Class 12 - Maths - Differential Equations

Exercise 9.1

Question 1:

Determine order and degree (if defined) of differential equation d4y/dy4 + sin (y”) = 0

Given, d4y/dy4 + sin (y”) = 0

=> y”” + sin (y”) = 0

The highest order derivative present in the differential equation is y’’’’.

Therefore, its order is four.

The given differential equation is not a polynomial equation in its derivatives.

Hence, its degree is not defined.

Question 2:

Determine order and degree (if defined) of differential equation y’ + 5y = 0

The given differential equation is: y’ + 5y = 0

The highest order derivative present in the differential equation is y’.

Therefore, its order is one.

It is a polynomial equation in y’. The highest power raised to is 1.

Hence, its degree is one.

Question 3:

Determine order and degree (if defined) of differential equation (ds/dt)4 + 3s * d2s/dt2 = 0

Give, differential equation is: (ds/dt)4 + 3s * d2s/dt2 = 0

The highest order derivative present in the given differential equation is d2s/dt2.

Therefore, its order is two.

It is a polynomial equation in d2s/dt2 and ds/dt. The power raised to is 1.

Hence, its degree is one.

Question 4:

Determine order and degree (if defined) of differential equation (d2y/dt2)2 + cos (dy/dx) = 0

The highest order derivative present in the given differential equation is (d2y/dt2)2.

Therefore, its order is 2.

The given differential equation is not a polynomial equation in its derivatives.

Hence, its degree is not defined.

Question 5:

Determine order and degree (if defined) of differential equation d2y/dt2 = cos 3x + sin 3x

Given, differential equation is: d2y/dt2 = cos 3x + sin 3x

=> d2y/dt2 - cos 3x - sin 3x = 0

The highest order derivative present in the differential equation is d2y/dt2.

Therefore, its order is two.

It is a polynomial equation in d2y/dt2 and the power raised to d2y/dt2 is 1.

Hence, its degree is one.

Question 6:

Determine order and degree (if defined) of differential equation

(y”’)2 + (y”)3 + (y’)4 + y5 = 0

Given, differential equation is: (y”’)2 + (y”)3 + (y’)4 + y5 = 0

The highest order derivative present in the differential equation is y’’’.

Therefore, its order is three.

The given differential equation is a polynomial equation in y’’’, y’’ and y’

The highest power raised to y’’’ is 2.

Hence, its degree is 2.

Question 7:

Determine order and degree (if defined) of differential equation y’’’ + 2y’’ + y’ = 0

Given, differential equation is: y’’’ + 2y’’ + y’ = 0

The highest order derivative present in the differential equation is y’’’.

Therefore, its order is three.

It is a polynomial equation in y’’’, y’’ and y’. The highest power raised to y’’’ is 1.

Hence, its degree is 1.

Question 8:

Determine order and degree (if defined) of differential equation y’ + y = ex

Given, differential equation is: y’ + y = ex

=> y’ + y - ex = 0

The highest order derivative present in the differential equation is y’.

Therefore, its order is one.

The given differential equation is a polynomial equation in y’ and the highest power raised to

y’ is one. Hence, its degree is one.

Question 9:

Determine order and degree (if defined) of differential equation y” + (y’)2 + 2y = 0

Given, differential equation is: y” + (y’)2 + 2y = 0

The highest order derivative present in the differential equation is y”.

Therefore, its order is two.

The given differential equation is a polynomial equation in y” and y’ the highest power raised

to y” is one. Hence, its degree is one.

Question 10:

Determine order and degree (if defined) of differential equation y” + 2y’ + sin y = 0

Given, differential equation is: y” + 2y’ + sin y = 0

The highest order derivative present in the differential equation is y”.

Therefore, its order is two.

This is a polynomial equation in y” and y’ the highest power raised to y” is one.

Hence, its degree is one.

Question 11:

The degree of the differential equation (d2y/dx2)2 + (dy/dx)2 + sin(dy/dx) + 1 = 0 is                     (A) 3                              (B) 2                                (C) 1                     (D) not defined

Given, differential equation is: (d2y/dx2)2 + (dy/dx)2 + sin(dy/dx) + 1 = 0

The given differential equation is not a polynomial equation in its derivatives.

Therefore, its degree is not defined.

Hence, the correct answer is option D.

Question 12:

The order of the differential equation is 2x2 * d2y/dx2 - 3dy/dx + y = 0                                               (A) 2                               (B) 1                             (C) 0                          (D) not defined

Given, differential equation is: 2x2 * d2y/dx2 - 3dy/dx + y = 0

The highest order derivative present in the given differential equation is d2y/dx2.

Therefore, its order is two.

Hence, the correct answer is option A.

Exercise 9.2

In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

Question 1:

y = ex + 1             :              y’’ – y’ = 0

Given, y = ex + 1

Differentiating both sides of this equation with respect to x, we get:

dy/dx = d(ex + 1)/dx

=> y’ = ex   …………1

Now, differentiating equation 1 with respect to x, we get:

d(y’)/dx = d(ex)/dx

=> y” = ex   …………2

Substituting the values of in the given differential equation, we get the L.H.S. as:

y’’ – y’ = ex - ex = 0 = R.H.S.

Thus, the given function is the solution of the corresponding differential equation.

Question 2:

y = x2 + 2x + C         :                 y’ – 2x – 2 = 0

Given, equation is: y = x2 + 2x + C

Differentiating both sides of this equation with respect to x, we get:

y’= d(x2 + 2x + C)/dx

=> y’=  2x + 2

Put the value of y’ in the given differential equation, we get:

L.H.S. = (2x + 2) – 2x – 2 = 2x + 2 – 2x – 2 = 0 = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Question 3:

y = cos x + C            :                  y’ + sin x = 0

Given, y = cos x + C

Differentiating both sides of this equation with respect to x, we get:

y’ = d(cos x + C)/dx

=> y’ = -sin x

Put the value of y’ in the given differential equation, we get:

L.H.S. = y’ + sin x = -sin x + sin x = 0 = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Question 4:

y = √(1 + x2)                 :                 y’ = xy/(1 + x2)

Given, y = √(1 + x2)

Differentiating both sides of the equation with respect to x, we get:

y’ = d(√(1 + x2))/dx

=> y’ = {1/2√(1 + x2)} * d(1 + x2)/dx

=> y’ = 2x/2√(1 + x2)

=> y’ = x/√(1 + x2)

=> y’ = {x/√(1 + x2)} * {√ (1 + x2)/√(1 + x2)}

=> y’ = x√(1 + x2)/(1 + x2)

=> y’ = xy/(1 + x2)                        [Since y = √(1 + x2)]

So, L.H.S. = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Question 5:

y = Ax                 :                       xy’ = y (x ≠ 0)

Given, y = Ax

Differentiating both sides with respect to x, we get:

y’ = A

Put the value of y’ in the given differential equation, we get:

L.H.S. = xy’ = x * A = Ax = y = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Question 6:

y = x sin x              :                     xy’ = y + x√(x2 – y2)     (x ≠ 0 and x > y or x < -y)

Given, y = x sin x

Differentiating both sides of this equation with respect to x, we get:

y’ = sin x * d(x)/dx + x * d(sin x)/dx

=> y’= sin x + x cos x

Put the value of y’ in the given differential equation, we get:

LHS = xy’ = x(sin x + x cos x)

= x sin x + x2 cos x

= x sin x + x2 √(1 – sin2 x)

= x sin x + x2 √{1 – (y/x)2}

= x sin x + x √(x2 – y2)

= y + x √(x2 – y2

= R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Question 7:

xy = log y + C          :                  y’ = y2/(1 - xy)   (xy ≠ 1)

Given, xy = log y + C

Differentiating both sides of this equation with respect to x, we get:

d(xy)/dx = d(log y + C)/dx

=> y * d(x)/dx + x * dy/dx = (1/y) * dy/dx

=> y + xy’ = y’/y

=> y2 + xy y’ = y’

=> y2 = y’ – xy y’

=> y2 = y’(1 – xy)

=> y’ = y2 /(1 - xy)

So, L.H.S. = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Question 8:

y – cos y = x                :                   (y sin y + cos y + x)y’ = y

Given, y – cos y = x  …………..1

Differentiating both sides of the equation with respect to x, we get:

dy/dx – d(cos y)/dx = d(x)/dx

=> y’ + sin y * y’ = 1

=> y’(1 + sin y) = 1

=> y’ = 1/(1 + sin y)

Put the value of y’ in the given differential equation, we get:

L.H.S.:

(y sin y + cos y + x)y’ = (y sin y + cos y – cos y) * {1/(1 + sin y)}

= y(1 +  sin y) * {1/(1 + sin y)}

= y

= R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Question 9:

x + y = tan-1 y     :    y2y’ + y2 + 1 = 0

Given, x + y = tan-1 y

Differentiating both sides of this equation with respect to x, we get:

d(x + y)/dx = d(tan-1 y)/dx

=> 1 + y’ = {1/(1 + y2)} * y’

=> y’{1/(1 + y2) - 1} = 1

=> y’[{1 - (1 + y2)}/(1 + y2)] = 1

=> y’[-y2/(1 + y2)] = 1

=> y’ = -(1 + y2)/y2

Substituting the value of y’ in the given differential equation, we get

L.H.S.:

y2y’ + y2 + 1 = y2[-(1 + y2)/y2] + y2 + 1

= -1 - y2 + y2 + 1

= 0

= R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Question 10:

y = √(a2 – x2), x є (-a, a)    :              x + y * dy/dx = 0, (y ≠ 0)

Given, y = √(a2 – x2)

Differentiating both sides of this equation with respect to x, we get:

dy/dx = d√(a2 – x2)/dx

=> dy/dx = 1/2√(a2 – x2) * d(a2 – x2)/dx

=> dy/dx = 1/2√(a2 – x2) * (-2x)

=> dy/dx = -x/√(a2 – x2)

Put the value of dy/dx, we get

L.H.S.:

x + y * dy/dx = x + √(a2 – x2) * {-x/√(a2 – x2)}

= x – x

= 0

= R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Question 11:

The numbers of arbitrary constants in the general solution of a differential equation of fourth order are:                                                                                                                                                     (A) 0                             (B) 2                                (C) 3                                (D) 4

We know that the number of constants in the general solution of a differential equation of

order n is equal to its order.

Therefore, the number of constants in the general equation of fourth order differential

equation is four.

Hence, the correct answer is option D.

Question 12:

The numbers of arbitrary constants in the particular solution of a differential equation of third order are:                                                                                                                                                     (A) 3                               (B) 2                            (C) 1                                      (D) 0

In a particular solution of a differential equation, there are no arbitrary constants.

Hence, the correct answer is option D.

Exercise 9.3

In each of the Exercises 1 to 5, form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

Question 1:

x/a + y/b = 1

Given, x/a + y/b = 1

Differentiating both sides of the given equation with respect to x, we get:

1/a + (1/b) * dy/dx = 0

=> 1/a + (1/b) * y’ = 0

Again, differentiating both sides with respect to x, we get:

0 + (1/b) * y” = 0

=> y” = 0

Hence, the required differential equation of the given curve is y” = 0

Question 2:

y2 = a(b2 – x2)

Given, y2 = a(b2 – x2)

Differentiating both sides with respect to x, we get:

2y * dy/dx = -2ax

=> yy’ = -2ax   …………1

Again, differentiating both sides with respect to x, we get:

y’ * y’ + yy” = -2a

=> (y’)2 + yy” = -2a  ……….2

Dividing equation 2 by equation 1, we get:

{(y’)2 + yy”}/yy’ = -a/-ax

=> xyy” + x(y’)2 – yy” = 0

This is the required differential equation of the given curve.

Question 3:

y = ae3x + be-2x

Given, y = ae3x + be-2x   …………1

Differentiating both sides with respect to x, we get:

y’ = 3ae3x - 2be-2x   ………………2

Again, differentiating both sides with respect to x, we get:

y’’ = 9ae3x + 4be-3x   ………………3

Multiplying equation 1 by 2 and then adding it to equation 2, we get:

(2ae3x + 2be-2x) + (3ae3x - 2be-2x) = 2y + y’

=> 5ae3x = 2y + y’

=> ae3x = (2y + y’)/5

Now, multiplying equation 1 with equation 3 and subtracting equation 2 from it, we get:

(3ae3x + 3be-2x) - (3ae3x - 2be-2x) = 3y – y’

=> 5be3x = 3y - y’

=> be3x = (3y - y’)/5

Substituting the values of ae3x and be-2x in equation (3), we get:

y” = 9 * {(2y + y’)/5} + 4 * {(3y - y’)/5}

=> y” = (18y + 9y’)/5 + (12y -4 y’)/5

=> y” = (30y + 5y’)/5

=> y” = 6y + y’

=> y” - 6y - y’ = 0

This is the required differential equation of the given curve.

Question 4:

y = e2x(a + bx)

Given, y = e2x(a + bx)   ………….1

Differentiating both sides with respect to x, we get:

y’ = 2e2x(a + bx) + e2x * b

=> y’ = e2x(2a + 2bx + b)   ………2

Multiplying equation 1 with equation 2 and then subtracting it from equation 2, we get:

y’ – 2y = e2x(2a + 2bx + b) - e2x(2a + 2bx)

=> y’ – 2y = be2x  ……………3

Differentiating both sides with respect to x, we get:

=> y” – 2y’ = 2be2x  ……….4

Dividing equation 4 by equation 3, we get:

(y” – 2y’)/(y’ – 2y) = 2be2x / be2x

=> (y” – 2y’)/(y’ – 2y) = 2

=> (y” – 2y’) = 2(y’ – 2y)

=> (y” – 2y’) = 2y’ – 4y

=> y” – 4y’ + 4y = 0

This is the required differential equation of the given curve.

Question 5:

y = ex(a cos x + b sin x)

Given, y = ex(a cos x + b sin x)

Differentiating both sides with respect to x, we get:

y’ = ex(a cos x + b sin x) + ex(-a sin x + b cos x)

=> y’ = ex[(a + b)cos x – (a – b) sin x]  ……….2

Again, differentiating with respect to x, we get:

y’’ = ex[(a + b)cos x – (a – b) sin x] + ex[-(a + b)sin x – (a – b) cos x]

=> y’’ = ex[2b cos x – 2a sin x]

=> y’’ = 2ex[b cos x – a sin x]

=> y’’/2 = ex[b cos x – a sin x]   ……………..3

Adding equations 1 and 2, we get:

y + y”/2 = ex[(a + b) cos x – (a - b) sin x]

=> y + y”/2 = y’

=> 2y + y” = 2y’

=> y” – 2y’ + 2y = 0

This is the required differential equation of the given curve.

Question 6:

Form the differential equation of the family of circles touching the y-axis at the origin.

The centre of the circle touching the y-axis at origin lies on the x-axis.

Let (a, 0) be the centre of the circle.

Since it touches the y-axis at origin, its radius is a.

Now, the equation of the circle with centre (a, 0) and radius (a) is

(x - a)2 + y2 = a2

=> x2 + y2 = 2ax   ………..1

Differentiating equation 1 with respect to x, we get:

2x + 2yy’ = 2a

=> x + yy’ = a

Now, on substituting the value of a in equation 1, we get:

=> x2 + y2 = 2(x + yy’)x

=> x2 + y2 = 2x2 + 2xyy’

=> 2xyy’ + x2 = y2

This is the required differential equation.

Question 7:

Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.

The equation of the parabola having the vertex at origin and the axis along the positive y-axis

is given as:

x2 = 4ay   …………1

Differentiating equation 1 with respect to x, we get:

2x = 4ay’   ……….2

Dividing equation 2 by equation 1, we get:

2x/x2 = 4ay’/4ay

=> 2/x = y’/y

=> xy’ = 2y

=> xy’ – 2y = 0

This is the required differential equation.

Question 8:

Form the differential equation of the family of ellipses having foci on y-axis and centre at origin.

The equation of the family of ellipses having foci on the y-axis and the centre at origin is as

follows:

x2/b2 + y2/a2 = 1   …………….1

Differentiating equation 1 with respect to x, we get:

2x/b2 + 2yy’/a2 = 0

=> x/b2 + yy’/a2 = 0

Again, differentiating with respect to x, we get:

=> 1/b2 + (y’.y’ + y.y”)/a2 = 0

=> 1/b2 + (y’2 + y.y”)/a2 = 0

=> 1/b2 = -(y’2 + y.y”)/a2

Substituting this value in equation 2, we get:

x[-(y’2/a2 + yy”)] + y.y’/a2 = 0

=> - xy’2 - xyy” + yy’/a2 = 0

=> xyy” + xy’2  - yy’/a2 = 0

This is the required differential equation.

Question 9:

Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.

The equation of the family of hyperbolas having foci on x-axis and centre at the origin is:

x2/a2 - y2/b2 = 1   …………….1

Differentiating both sides of equation 1 with respect to x, we get:

2x/a2 – 2yy’/b2 = 0

=> x/a2 – yy’/b2 = 0    …………..2

Again, differentiating both sides with respect to x, we get:

=> 1/a2 - (y’.y’ + y.y”)/b2 = 0

=> 1/a2 = (y’2 + yy”)/b2

Put the value of 1/a2 in equation 2, we get

x[-(y’2 + yy”)]/b2 - y.y’/b2 = 0

=> - xy’2 + xyy” - yy’ = 0

=> xyy” + xy’2  - yy’/a2 = 0

This is the required differential equation.

Question 10:

Form the differential equation of the family of circles having centres on y-axis and radius 3 units.

Let the center of the circle on y-axis be (0, b).

The differential equation of the family of circles with centre at (0, b) and radius 3 is as follows:

x2 + (y - b)2 = 32

=> x2 + (y - b)2 = 9   …………….1

Differentiate equation 1 w.r.t. x, we get

=> 2x + 2(y – b).y’ = 0

=> (y – b).y’ = -x

=> y – b = -x/y’

Put the value of (y - b) in equation 1, we get

=> x2 + (-x/y’)2 = 9

=> x2 [1 + 1/y’2 ]= 9

=> x2 [y’2 + 1]= 9y’2

=> x2 [y’2 + 1]= 9y’2

=> (x2 - 9) y’2 + x2 = 0

This is the required differential equation.

Question 11:

Which of the following differential equations has y = c1ex + c2e-x as the general equation

A. d2y/dx2 + y = 0           B. d2y/dx2 - y = 0          C. d2y/dx2 + 1 = 0          D. d2y/dx2 - 1 = 0

The given equation is: y = c1ex + c2e-x  ………….1

Differentiating with respect to x, we get:

dy/dx = c1ex - c2e-x

Again, differentiating with respect to x, we get:

=> dy2/dx2 = c1ex + c2e-x

=> dy2/dx2 = y

=> dy2/dx2 – y = 0

This is the required differential equation of the given equation of curve.

Hence, the correct answer is option B.

Question 12:

Which of the following differential equation has y = x as one of its particular solution?

1. d2y/dx2 – x2 * dy/dx + xy = x B. d2y/dx2 + x * dy/dx + xy = x
2. d2y/dx2 – x2 * dy/dx + xy = 0 D. d2y/dx2 + x * dy/dx + xy = 0

Given, equation of curve is: y = x

Differentiate w.r.t. x, we get

dy/dx = 1   ……………1

Again, differentiate w.r.t. x, we get

d2y/dx2 = 0 ………….2

Now, on substituting the value of d2y/dx2 and dy/dx from equation 1 and 2 in each of the given

alternatives, we find that only the differential equation given in alternative C is correct.

d2y/dx2 – x2 * dy/dx + xy = 0 – x2 * 1 + x.x

= -x2 + x2

= 0

Hence, the correct answer is option C.

Exercise 9.4

For each of the differential equations in Exercises 1 to 10, find the general solution:

Question 1:

dy/dx = (1 – cos x)/(1 + cos x)

The given differential equation is:

dy/dx = (1 – cos x)/(1 + cos x)

=> dy/dx = (2sin2 x/2)/ (2cos2 x/2)

=> dy/dx = tan2 x/2

=> dy/dx = sec2 x/2 – 1

Separating the variables, we get

=> dy = (sec2 x/2 – 1)dx

Now, integrating both sides of this equation, we get:

=> ꭍ dy = ꭍ (sec2 x/2 – 1)dx

=> ꭍ dy = ꭍ sec2 x/2 dx - ꭍ dx

=> y = 2 tan x/2 – x + C

This is the required general solution of the given differential equation.

Question 2:

dy/dx = √(4 – y2), (-2 < y < 2)

The given differential equation is:

dy/dx = √(4 – y2)

Separating the variables, we get

dy = √(4 – y2) dx

Now, integrating both sides of this equation, we get:

ꭍ dy = ꭍ √(4 – y2) dx

=> sin-1 y/2 = x + C

=> y/2 = sin(x + C)

=> y = 2 sin(x + C)

This is the required general solution of the given differential equation.

Question 3:

dy/dx + y = 1, (y ≠ 1)

The given differential equation is:

dy/dx + y = 1

=> dy + y dx = dx

=> dy = dx – y dx

=> dy = (1 - y)dx

Seperating the variables, we get

=> dy/(1 - y) = dx

Now, integrating both sides of this equation, we get:

=> ꭍ dy/(1 - y) = ꭍ dx

=> log(1 – y) = x + log C

=> log(1 – y) - log C = x

=> log{C(1 - y)} = x

=> C(1 - y) = ex

=> (1 - y) = ex/C

=> y = 1 - ex/C

=> y = 1 + Aex, where A = -1/C

This is the required general solution of the given differential equation.

Question 4:

sec2 x tan y dx + sec2 y tan x dy = 0

The given differential equation is:

sec2 x tan y dx + sec2 y tan x dy = 0

=> (sec2 x tan y dx + sec2 y tan x dy)/(tan x * tan y) = 0/(tan x * tan y)

=> sec2 x/tan x dx + sec2 y/tan y dy = 0

=> sec2 x/tan x dx = -sec2 y/tan y dy

Integrating both sides of this equation, we get:

=> ꭍ sec2 x/tan x dx = - ꭍ sec2 y/tan y dy   ………….1

Let tan x = t

=> d(tan x)/dx = dt/dx

=> sec2 x = dt/dx

=> sec2 x dx = dt

Now, ꭍ sec2 x/tan x dx = ꭍ 1/t dt

= log t

= log (tan x)

Similarly, ꭍ sec2 x/tan x dx = log (tan y)

Substituting these values in equation 1, we get:

log (tan x) = - log (tan y) + log C

=> log (tan x) = log (C/tan y)

=> tan x= C/tan y

=> tan x * tan y = C

This is the required general solution of the given differential equation.

Question 5:

(ex + e-x)dy - (ex - e-x)dx = 0

The given differential equation is:

(ex + e-x)dy - (ex - e-x)dx = 0

=> (ex + e-x)dy = (ex - e-x)dx

=> dy = [(ex - e-x)/(ex + e-x)]dx

Integrating both sides of this equation, we get:

=> ꭍ dy = ꭍ [(ex - e-x)/(ex + e-x)]dx

=> y = ꭍ [(ex - e-x)/(ex + e-x)]dx  …………1

Let (ex + e-x) = t

Differentiating both sides with respect to x, we get:

=> d(ex + e-x)/dx = dt/dx

=> ex - e-x = dt/dx

=> (ex - e-x)dx = dt

Substituting this value in equation (1), we get:

=> y = ꭍ 1/t dt + C

=> y = log t + C

=> y = log (ex + e-x) + C

This is the required general solution of the given differential equation.

Question 6:

dy/dx = (1 + x2)(1 + y2)

The given differential equation is:

dy/dx = (1 + x2)(1 + y2)

=> dy/(1 + y2) = (1 + x2) dx

Integrating both sides of this equation, we get:

=> ꭍ dy/(1 + y2) = ꭍ (1 + x2) dx

=> tan-1 x = ꭍ dx + ꭍ x2 dx

=> tan-1 x = x + x3/3 + C

This is the required general solution of the given differential equation.

Question 7:

y log y dx – x dy = 0

The given differential equation is:

y log y dx – x dy = 0

=> y log y dx = x dy

=> dy/(y log y) = dx/x

Integrating both sides, we get:

=> ꭍ dy/(y log y) = ꭍ dx/x   ………….1

Let log y = t

=> d(log y)/dy = dt/dy

=> 1/y = dt/dy

=> dy/y = dt

Substituting this value in equation 1, we get:

=> ꭍ dt/t = ꭍ dx/x

=> log t = log x + log C

=> log (log y) = log Cx

=> log y = Cx

=> y = eCx

This is the required general solution of the given differential equation.

Question 8:

x5 * dy/dx = -y5

The given differential equation is:

dy/dx = -y5

=> dy/y5 = - dx/x5

=> dy/y5 + dx/x5 = 0

Integrating both sides, we get:

=> ꭍ dy/y5 + ꭍ dx/x5 = k, where k is any constant

=> ꭍ y-5 dy + ꭍ x-5 dx = k

=> y-4/(-4) + x-4/(-4) = k

=> y-4 + x-4 = -4k

=> x-4 + y-4 = C, where C = -4k

This is the required general solution of the given differential equation.

Question 9:

dy/dx = sin-1 x

The given differential equation is:

dy/dx = sin-1 x

=> dy = sin-1 x dx

Integrating both sides, we get:

=> ꭍ dy = ꭍ sin-1 x dx

=> y = ꭍ (sin-1 x * 1) dx

=> y = sin-1 x * ꭍ 1 dx - ꭍ[d(sin-1 x)/dx * ꭍ 1 dx]dx

=> y = sin-1 x * x - ꭍ[1/√(1 – x2) * x]dx

=> y = x sin-1 x + ꭍ[-x/√(1 – x2)]dx   …………1

Let 1 – x2 = t

=> d(1 – x2)/dx = dt/dx

=> -2x = dt/dx

=> -xdx = dt/2

Substituting this value in equation 1, we get:

=> y = x sin-1 x + ꭍ 1/2√t dt

=> y = x sin-1 x + (1/2) * t1/2/(1/2)  + C

=> y = x sin-1 x + √t + C

=> y = x sin-1 x + √(1 – x2) + C

This is the required general solution of the given differential equation.

Question 10:

ex tan y dx + (1 - ex) sec2 y dy = 0

The given differential equation is:

ex tan y dx + (1 - ex) sec2 y dy = 0

=> (1 - ex) sec2 y dy = - ex tan y dx

Separating the variables, we get

=> sec2 y/tan y dy = -ex/(1 - ex) dx

Integrating both sides, we get:

=> ꭍ sec2 y/tan y dy = ꭍ -ex/(1 - ex) dx   …………..1

Let tan y = u

=> d(tan y)/dy = du/dy

=> sec2 y = du/dt

So, ꭍ sec2 y/tan y dy = ꭍ du/u = log u = log(tan y)

Again, let 1 - ex = t

=> d(1 - ex)/dx = dt/dx

=> - ex = dt/dx

=> - ex dx = dt

Now, ꭍ -ex/(1 - ex) dx = log t = log (1 - ex)

Substitute the values of ꭍ sec2 y/tan y dy and ꭍ -ex/(1 - ex) dx in equation 1, we get

=> log(tan y) = log (1 - ex) + log C

=> log(tan y) = log [C(1 - ex)]

=> tan y = C(1 - ex)

This is the required general solution of the given differential equation.

Question 11:

(x5 + x2 + x + 1) * dy/dx = 2x2 + x, y = 1 when x = 0

The given differential equation is:

(x5 + x2 + x + 1) * dy/dx = 2x2 + x

=> dy/dx = (2x2 + x)/ (x5 + x2 + x + 1)

=> dy/dx = (2x2 + x)/{(x + 1)(x2 + 1)}

=> dy = (2x2 + x)/{(x + 1)(x2 + 1)} dx

Integrating both sides, we get:

=> ꭍ dy = ꭍ (2x2 + x)/{(x + 1)(x2 + 1)} dx   …………1

Let (2x2 + x)/{(x + 1)(x2 + 1)} = A/(x + 1) + (Bx + C)/(x2 + 1)   ………….2

=> (2x2 + x)/{(x + 1)(x2 + 1)} = {A(x2 + 1) + (Bx + C)(x + 1)}/{ (x + 1)(x2 + 1)}

=> 2x2 + x = Ax2 + A + Bx2 + Bx + Cx + C

=> 2x2 + x = (A + B)x2 + (B + C)x + (A + C)

Comparing the coefficients of x2 and x, we get:

A + B = 2

B + C = 1

A + C = 0

Solving these equations, we get:

A = 1/2, B = 3/2 and C = -1/2

Substituting the values of A, B, and C in equation 2, we get:

(2x2 + x)/{(x + 1)(x2 + 1)} = 1/2(x + 1) + (3x - 1)/2(x2 + 1)

Therefore, equation 1 becomes

=> ꭍ dy = (1/2)ꭍ dx/(x + 1) + (3/2) ꭍ x/(x2 + 1) dx -  (1/2) ꭍ 1/(x2 + 1) dx

=> y = (1/2) log(x + 1) + (3/4) ꭍ 2x/(x2 + 1) dx - (1/2) ꭍ 1/(x2 + 1) dx + C

=> y = (1/2) log(x + 1) + (3/4) log(x2 + 1) - (1/2) tan-1 x + C

=> y = (1/4) [2 log(x + 1) + 3 log(x2 + 1)] - (1/2) tan-1 x + C

=> y = (1/4) [log {(x + 1)2 * (x2 + 1)3}] - (1/2) tan-1 x + C   ………..3

Now, y = 1 when x = 0

=> 1 = (1/4) [log 1] - (1/2) tan-1 0 + C

=> 1 = (1/4) * 0 - (1/2) * 0 + C

=> C = 1

Substituting C = 1 in equation 3, we get:

y = (1/4) [log {(x + 1)2 * (x2 + 1)3}] - (1/2) tan-1 x + 1

This is the required general solution of the given differential equation.

Question 12:

x(x2 - 1) * dy/dx = 1, y = 0 when x = 2

The given differential equation is:

x(x2 - 1) * dy/dx = 1

=> dy = dx/x(x2 - 1)

=> dy = dx/{x(x - 1)(x + 1)}

Integrating both sides, we get

=> ꭍ dy = ꭍ dx/{x(x - 1)(x + 1)}   …………….1

Let 1/{x(x - 1)(x + 1)} = A/x + B/(x - 1) + C/(x + 1)   ………..2

=> 1/{x(x - 1)(x + 1)} = {A(x - 1)(x + 1) + Bx(x + 1) + Cx(x - 1)}/{x(x - 1)(x + 1)}

=> 1 = A(x2 - 1) + B(x2 + x) + C(x2 - x)

=> 1 = (A + B + C)x2 + (B - C)x – A

Comparing the coefficients of x2, x, and constant, we get:

A = -1

B – C = 0

A + B + C = 0

Solving these equations, we get

A = -1, B = 1/2 and C = 1/2

Substituting the values of A, B, and C in equation 2, we get:

1/{x(x - 1)(x + 1)} = -1/x + 1/2(x - 1) + 1/2(x + 1)

Therefore, equation 1 becomes:

ꭍ dy = -ꭍ 1/x dx + (1/2) ꭍ (x - 1) dx + (1/2) ꭍ (x + 1) dx

=> y = -log x + (1/2) log (x - 1 + (1/2) log (x + 1) + log k

=> y = (1/2)log[{k2(x - 1)(x + 1)}/x2]   ………3

Now, y = 0 when x = 2

Substituting the value of k2 in equation 3, we get:

=> 0 = (1/2)log[{k2(2 - 1)(2 + 1)/4}]

=> log[3k2/4] = 0

=> 3k2/4 = 1

=> 3k2 = 4

=> k2 = 4/3

Substituting the value of k2 in equation 3, we get:

=> y = (1/2)log[{4(x - 1)(x + 1)}/3x2]

=> y = (1/2)log[4(x2 - 1)/3x2]

Question 13:

cos(dy/dx) = a (a є R); y = 1 when x = 0

The given differential equation is:

cos(dy/dx) = a

=> dy/dx = cos-1 a

=> dy = cos-1 a dx

Integrating both sides, we get:

=> ꭍ dy = cos-1 a ꭍ dx

=> y = cos-1 a * x + C

=> y = x cos-1 a + C  ………….1

Now, y = 1 when x = 0

=> 1 = 0 * cos-1 a + C

=> C = 1

Substituting C = 1 in equation 1, we get:

=> y = x cos-1 a + 1

=> y – 1 = x cos-1 a

=> (y – 1)/x = cos-1 a

=> cos(y – 1)/x = a

Question 14:

dy/dx = y tan x; y = 1 when x = 0

The given differential equation is:

dy/dx = y tan x

=> dy/y = tan x dx

Integrating both sides, we get

=> ꭍ dy/y = ꭍ tan x dx

=> log y = log sec x + log C

=> log y = log (C * sec x)

=> y = C * sec x     …………….1

Now, y = 1 when x = 0

=> 1 = C * sec 0

=> 1 = C * 1

=> C = 1

Put the value of C in equation 1, we get

y = sec x

Question 15:

Find the equation of a curve passing through the point (0, 0) and whose differential equation is y′ = ex sin x.

The differential equation of the curve is:

y′ = ex sin x

=> dy/dx = ex sin x

=> dy = ex sin x dx

Integrating on both sides, we get

=> ꭍ dy = ꭍ ex sin x dx

=> y = ꭍ ex sin x dx   ……..1

=> y = sin x ꭍ ex dx - ꭍ[d(sin x)/dx * ꭍ ex dx]dx

=> y = sin x * ex - ꭍ [cos x * ex]dx

=> y = sin x * ex – [cos x ꭍ ex dx - ꭍ{d(cos x)/dx * ꭍ ex dx}dx]

=> y = sin x * ex – [cos x * ex - ꭍ{ (-sin x) * ex}dx]

=> y = sin x * ex – cos x * ex - ꭍ sin x * ex dx

=> y = sin x * ex – cos x * ex – y

=> 2y = ex [sin x – cos x]

=> y = ex [sin x – cos x]/2 + C  ………….2

Now, curve passing through the point (0, 0)

So, 0 = e0 [sin 0 – cos 0]/2 + C

=> C = 1/2

Put the value of C in equation 2, we get

=> y = ex [sin x – cos x]/2 + 1/2

=> 2y = ex [sin x – cos x] + 1

=> 2y – 1 = ex [sin x – cos x]

Hence, the required equation of the curve is: 2y – 1 = ex [sin x – cos x]

Question 16:

For the differential equation xy * dy/dx = (x + 2)(y + 2), find the solution curve passing through the point (1, –1).

The differential equation of the curve is:

xy * dy/dx = (x + 2)(y + 2)

=> y/(y + 2) dy = (x + 2)/x dx

=> [1 - 2/(y + 2)] dy = (1 + 2/x) dx

Integrating on both sides, we get

=> ꭍ [1 - 2/(y + 2)] dy = ꭍ (1 + 2/x) dx

=> ꭍ [1 - 2ꭍ dy/(y + 2) dy = ꭍ dx + 2ꭍ 1/x dx

=> y – 2 log(y + 2) = x + 2 log x + C

=> y – x – C = log x2 + log(y + 2)2

=> y – x – C = log [x2(y + 2)2]  …………1

Now, the curve passes through the point (1, -1)

=> -1 – 1 – C = log [12(-1 + 2)2]

=> -2 – C = log 1

=> -2 – C = 0

=> C = -2

Put the value of C in equation 1, we get

=> y – x + 2 = log [x2(y + 2)2]

This is the required solution of the given curve.

Question 17:

Find the equation of a curve passing through the point (0, –2) given that at any point (x, y) on the curve,

the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

Let x and y be the x-coordinate and y-coordinate of the curve respectively.

We know that the slope of a tangent to the curve in the coordinate axis is given by the relation

dy/dx.

According to the given information, we get

y * dy/dx = x

=> y dy = x dx

Integrating on both sides, we get

=> ꭍ y dy = ꭍ x dx

=> y2/2 = x2/2 + C

=> y2 - x2 = 2C  ……..1

Now, the curve passes through the point (0, -2)

=> (-2)2 - 02 = 2C

=> 4 = 2C

=> 2C = 4

Put 2C = 4 in equation 1, we get

=> y2 - x2 = 4

This is the required equation of the curve.

Question 18:

At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3).

Find the equation of the curve given that it passes through (–2, 1).

It is given that (x, y) is the point of contact of the curve and its tangent.

The slope (m1) of the tangent to the curve is given by the relation

So, slope (m2) of the tangent = dy/dx

According to given information:

m2 = 2m1

=> dy/dx = 2(y + 3)/(x + 4)

=> dy/(y + 3) = 2dy/(x + 4)

Integrating on both sides, we get

=> ꭍ dy/(y + 3) = 2ꭍ dy/(x + 4)

=> log (y + 3) = 2log(x + 4) + log C

=> log (y + 3) = log(x + 4)2 + log C

=> y + 3 = C(x + 4)2    ………..1

This is the general equation of the curve.

It is given that the curve passes through the point (-2, 1)

=> 1 + 3 = C(-2 + 4)2

=> 4C = 4

=> C = 1

Put C = 1 in equation 1, we get

=> y + 3 = (x + 4)2

This is the required equation of the curve.

Question 19:

The volume of spherical balloon being inflated changes at a constant rate.

If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Let the rate of change of the volume of the balloon be k (where k is a constant).

=> dv/dt = k

=> d(4πr3/3)/dt = k

=> 4π/3 * 3r2 * dr/dt = k

=> 4π/3 * 3r2 * dr = k dt

=> 4πr2 * dr = k dt

Integrating on both sides, we get

=> 4π * ꭍ r2 dr = k ꭍ dt

=> 4πr3/3 = kt + C

=> 4πr3 = 3(kt + C)   ………….1

Now, at t = 0, r = 3

=> 4π * 33 = 3(k * 0 + C)

=> 108π = 3C

=> C = 36π

Again, at t = 3, r = 6

=> 4π * 63 = 3(k * 3 + C)

=> 864π = 3(3k + C)

=> 864π = 3(3k + 36π)

=> 288π = 3k + 36π

=> 3k = 288π - 36π

=> 3k = 252π

=> k = 84π

Put the value of k and C in equation 1, we get

=> 4πr3 = 3(84π t + 36π)

=> 4πr3 = 4π(63 t + 27)

=> r3 = (63 t + 27)

=> r3 = (63 t + 27)1/3

So, the radius of the balloon after t seconds is (63 t + 27)1/3

Question 20:

In a bank, principal increases continuously at the rate of r% per year.

Find the value of r if Rs 100 doubles itself in 10 years (loge 2 = 0.6931).

Let p, r and t represent the principal, time and rate of interest respectively.

It is given that the principal increases continuously at the rate of r% per year.

=> dp/dt = (r/100)p

=> dp/p = (r/100) dt

Integrating both sides, we get

=> ꭍ dp/p = ꭍ (r/100) dt

=> log p = rt/100 + k

=> p = ert/100 + k    ………….1

It is given that when t = 0, p = 100

=> 100 = ek

Now, if t = 10, then

=> p = 2 * 100 = 200

So, the equation 1 becomes

=> 200 = er/10 + k

=> 200 = er/10  * ek

=> 200 = er/10  * 100

=> er/10 = 2

=> r/10 = loge 2

=> r/10 = 0.6931

=> r = 6.931

=> r = 6.93 (approx.)

Hence, the value of r is 6.93%.

Question 21:

In a bank, principal increases continuously at the rate of 5% per year.

An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0.5 = 1.648).

Let p and t be the principal and time respectively.

It is given that the principal increases continuously at the time of 5% per year.

=> dp/dt = (5/100)p

=> dp/dt = p/20

=> dp/p = dt/20

Integrating both sides, we get

=> ꭍ dp/p = (1/20)ꭍ dt

=> log p = t/20 + C

=> p = et/20 + C    ………….1

It is given that when t = 0, p = 1000

=> 1000 = eC

At t = 10, equation 1 becomes

=> p = e1/2 + C

=> p = e0.5 * eC

=> p = 1.648  * 1000

=> p = 1648

Hence, after 10 years, the amount will worth Rs 1648.

Question 22:

In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours.

In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Let y be the number of bacteria at any instant t.

It is given that he rate of growth of the bacteria is proportional to the number present.

So, dy/dt is proportional to y

=> dy/dt = ky, where k is a constant

=> dy/y = k dt

Integrating on both sides, we get

=> ꭍ dy/y = k ꭍ dt

=> log y = kt + C  ……….1

Let y0 be the number of bacteria at t = 0

=> log y0 =  C

Put value of C in equation 1, we get

=> log y = kt + log y0

=> log y - log y0 = kt

=> log (y/y0) = kt

=> kt = log (y/y0)   ……….2

Again, it is given that the number of bacteria increase by 10% in hours.

So, y = (110/100) y0

=> y/ y0 = 110/100

=> y/ y0 = 11/10

Put this value in equation 2, we get

k * 2 = log (11/10)   ………3

=> k = (1/2) * log (11/10)

So, the equation 2 becomes

=> (1/2) * log (11/10) * t = log (y/y0)

=> t = {2 log (y/y0)}/ log (11/10)   …..4

Now, let the time when the number of bacteria increases from 100000 to 200000 be t1

=> y = 2y0 at t = t1

From equation 4, we get

=> t1 = {2 log (2 y0/y0)}/ log (11/10)

=> t1 = {2 log 2}/ log (11/10)

Hence, in {2 log 2}/ log (11/10) hours, the number of bacteria increase from 100000 to 200000

Question 23:

The general solution of the differential equation dy/dx = ex + y

(A) ex + e–y = C                (B) ex + ey = C                  (C) e-x + ey = C                  (D) e-x + e–y  = C

Given, dy/dx = ex + y

=> dy/dx = ex * ey

=> dy/ey = ex dx

=> e-y dy = ex dx

Integrating on both sides, we get

=> ꭍ e-y dy = ꭍ ex dx

=> -e-y = ex + k

=> ex + e-y = -k

=> ex + e-y = C, where C = -k

Hence, the correct answer is option A.

Exercise 9.5

In each of the Exercises 1 to 10, show that the given differential equation is homogeneous and solve each of them.

Question 1:

(x2 + xy)dy = (x2 + y2)dx

The given differential equation (x2 + xy)dy = (x2 + y2)dx can be written as:

dy/dx = (x2 + y2)/ (x2 + xy)   ………..1

Let F(x, y) = (x2 + y2)/(x2 + xy)

Now, F(kx, ky) = {(kx)2 + (ky)2}/{(kx)2 +(kx)(ky)}

=> F(kx, ky) = (x2 + y2)/(x2 + xy)

=> F(kx, ky) = k0 * F(x, y)

This shows that equation 1 is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Differentiating both sides with respect to x, we get:

dy/dx = v + x * dv/dx

Substituting the values of y and dy/dx in equation 1, we get:

v + x * dv/dx = {x2 + (vx)2}/{x2 + x(vx)}

=> v + x * dv/dx = (1 + v2)/(1 + v)

=> x * dv/dx = (1 + v2)/)1 + v) – v

=> x * dv/dx = {(1 + v2) – v(1 + v) }/(1 + v)

=> x * dv/dx = (1 + v2 – v - v2) }/(1 + v)

=> x * dv/dx = (1 - v)/(1 + v)

=> (1 + v)/(1 - v) dv = dx/x

=> {2/(1 + v) – 1} dv = dx/x

Integrating both sides, we get:

=> -2log(1 - v) – v = log x – log k

=> v = -2log(1 - v) – log x + log k

=> v = log k – [log x + log (1 - v)2]

=> v = log k – log [x(1 - v)2]

=> v = log[k/x(1 - v)2]

=> y/x = log[k/x(1 – y/x)2]

=> y/x = log[kx/(x – y)2]

=> kx/(x – y)2 = ey/x

=> (x – y)2 = kx * ey/x

This is the required solution of the given differential equation.

Question 2:

y’ = (x + y)/x

The given differential equation is:

y’ = (x + y)/x

=> dy/dx = (x + y)/x   ………..1

Let F(x, y) = (x + y)/x

Now, F(kx, ky) = (kx + ky)/kx

=> F(kx, ky) = (x + y)/x

=> F(kx, ky) = k0 * F(x, y)

This shows that equation 1 is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Differentiating both sides with respect to x, we get:

dy/dx = v + x * dv/dx

Substituting the values of y and dy/dx in equation 1, we get:

v + x * dv/dx = (x + vx)/x

=> v + x * dv/dx = 1 + v

=> x * dv/dx = 1

=> dv = dx/x

Integrating both sides, we get:

=> v = log x + C

=> y/x = log x + C

=> y = x log x + Cx

This is the required solution of the given differential equation.

Question 3:

(x - y)/dy – (x + y)dx = 0

The given differential equation is:

(x - y)/dy – (x + y)dx = 0

=> dy/dx = (x + y)/(x - y)   ………..1

Let F(x, y) = (x + y)/(x - y)

Now, F(kx, ky) = (kx + ky)/(kx - ky)

=> F(kx, ky) = (x + y)/(x - y)

=> F(kx, ky) = k0 * F(x, y)

This shows that equation 1 is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Differentiating both sides with respect to x, we get:

dy/dx = v + x * dv/dx

Substituting the values of y and dy/dx in equation 1, we get:

v + x * dv/dx = (x + vx)/ (x - vx)

=> v + x * dv/dx = (1 + v)/(1 - v)

=> x * dv/dx = (1 + v)/(1 - v) – v

=> x * dv/dx = {(1 + v) - (1 - v)v}/(1 - v)

=> x * dv/dx = (1 + v2)/(1 - v)

=> (1 - v)/(1 + v2) dv = dx/x

=> {1/(1 + v2) - v/(1 + v2)} = dx/x

Integrating both sides, we get:

=> tan-1 v – log (1 + v2)/2= log x + C

=> tan-1 (y/x) – log {1 + (y/x)2}/2= log x + C

=> tan-1 (y/x) – (1/2)[log {x2 + y2)/x2}] = log x + C

=> tan-1 (y/x) – (1/2)[log (x2 + y2) - log x2] = log x + C

=> tan-1 (y/x) = (1/2)[log (x2 + y2)] + C

This is the required solution of the given differential equation.

Question 4:

(x2 – y2)dx + 2xy dy = 0

The given differential equation is:

(x2 – y2)dx + 2xy dy = 0

=> dy/dx = -(x2 – y2)/2xy   ………..1

Let F(x, y) = -(x2 – y2)/2xy

Now, F(kx, ky) = -{(kx)2 – (ky)2}/{2(kx)(ky)}

=> F(kx, ky) = -(x2 – y2)/2xy

=> F(kx, ky) = k0 * F(x, y)

This shows that equation 1 is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Differentiating both sides with respect to x, we get:

dy/dx = v + x * dv/dx

Substituting the values of y and dy/dx in equation 1, we get:

v + x * dv/dx = -{x2 – (vx)2}/{2x(vx)}

=> v + x * dv/dx = (v2 - 1)/2v

=> x * dv/dx = [(v2 - 1)/2v] – v

=> x * dv/dx = (v2 - 1 - 2v2)/2v

=> x * dv/dx = -(1 + v2)/2v

=> 2v/(1 + v2) dv = -dx/x

Integrating both sides, we get:

=> log(1 + v2) = -log x + log C

=> log(1 + v2) = log (C/x)

=> 1 + v2 = C/x

=> 1 + (y/x)2 = C/x

=> (x2 + y2)/x2 = C/x

=> x2 + y2 = Cx

This is the required solution of the given differential equation.

Question 5:

x2 * dy/dx = x2 - 2y2 + xy = 0

The given differential equation is:

x2 * dy/dx = x2 - 2y2 + xy = 0

=> dy/dx = (x2 - 2y2 + xy)/x2   ………..1

Let F(x, y) = (x2 - 2y2 + xy)/x2

Now, F(kx, ky) = {(kx)2 – 2(ky)2 + (kx)(ky)}/(kx)2

=> F(kx, ky) = (x2 - 2y2 + xy)/x2

=> F(kx, ky) = k0 * F(x, y)

This shows that equation 1 is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Differentiating both sides with respect to x, we get:

dy/dx = v + x * dv/dx

Substituting the values of y and dy/dx in equation 1, we get:

v + x * dv/dx = {x2 – 2(vx)2 + x(vx)}/x2

=> v + x * dv/dx = 1 - 2v2 + v

=> x * dv/dx = 1 - 2v2

=> dv/(1 - 2v2) = dx/x

=> (1/2) * dv/(1/2 - v2) = dx/x

=> (1/2) * dv/{(1/√2)2 - v2} = dx/x

Integrating both sides, we get:

=> (1/2) * 1/(2 * 1/√2) * log|(1/√2 + v)/ (1/√2 - v)| = log |x| + C

=> 1/2√2 * log|(1/√2 + y/x)/ (1/√2 – y/x)| = log |x| + C

=> 1/2√2 * log|(x + √2y)/ (x – √2y)| = log |x| + C

This is the required solution of the given differential equation.

Question 6:

xdy - ydx = √(x2 + y2) dx

The given differential equation is:

xdy - ydx = √(x2 + y2) dx

=> xdy = [y + √(x2 + y2)] dx

=> dy/dx = [y + √(x2 + y2)]/x   ………..1

Let F(x, y) = [y + √(x2 + y2)]/x

Now, F(kx, ky) = [ky + √{(kx)2 + (ky)2}]/x

=> F(kx, ky) = [y + √(x2 + y2)]/x

=> F(kx, ky) = k0 * F(x, y)

This shows that equation 1 is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Differentiating both sides with respect to x, we get:

dy/dx = v + x * dv/dx

Substituting the values of y and dy/dx in equation 1, we get:

v + x * dv/dx = [vx + √{x2 + (vx)2}]/x

=> v + x * dv/dx = v + √(1 + v2)

=> x * dv/dx = √(1 + v2)

=> dv/√(1 + v2) = dx/x

Integrating both sides, we get:

=> log|v + √(1 + v2)| = log |x| + Log C

=> log|y/x + √{1 + (y/x)2}| = log |x| + log C

=> log|{y + √(x2 + y2)/x| = log |Cx|

=> {y + √(x2 + y2)}/x = Cx

=> y + √(x2 + y2) = Cx2

This is the required solution of the given differential equation.

Question 7:

{x * cos(y/x) + y * sin(y/x)}y dx = {y * sin(y/x) - x * cos(y/x)}x dx

The given differential equation is:

{x * cos(y/x) + y * sin(y/x)}y dx = {y * sin(y/x) - x * cos(y/x)}x dx

=> dy/dx = [{x * cos(y/x) + y * sin(y/x)}y]/[ {y * sin(y/x) - x * cos(y/x)}x]   …………1

Let F(x, y) = [{x * cos(y/x) + y * sin(y/x)}y]/[ {y * sin(y/x) - x * cos(y/x)}x]

Now, F(kx, ky) = [{kx * cos(ky/kx) + ky * sin(ky/kx)}ky]/[{ky * sin(ky/kx) - kx * cos(ky/kx)}kx]

=> F(kx, ky) = [{x * cos(y/x) + y * sin(y/x)}y]/[ {y * sin(y/x) - x * cos(y/x)}x]

=> F(kx, ky) = k0 * F(x, y)

This shows that equation 1 is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Differentiating both sides with respect to x, we get:

dy/dx = v + x * dv/dx

Substituting the values of y and dy/dx in equation 1, we get:

v + x * dv/dx = [{x * cos v + vx * sin v} * vx]/[{vx * sin v - x * cos v}x]

=> v + x * dv/dx = [v * cos v + v2 * sin v]/[v * sin v - cos v]

=> x * dv/dx = [v * cos v + v2 * sin v]/[v * sin v - cos v] – v

=> x * dv/dx = [(v * cos v + v2 * sin v) – (v2 * sin v + v * cos v)]/[v * sin v - cos v]

=> x * dv/dx = (2v * cos v)]/[v * sin v - cos v]

=> [v * sin v - cos v](v * cos v) dv = 2dx/x

=> (tan v – 1/v)dv = 2dx/x

Integrating both sides, we get:

=> log sec v – log v = 2log x + Log C

=> log (sec v/v) = log (Cx2)

=> sec v /v =Cx2

=> sec v = v * Cx2

=> sec (y/x) = (y/x) * Cx2

=> sec (y/x) = Cxy

=> cos (y/x) = 1/Cxy

=> cos (y/x) = 1/C * 1/xy

=> xy * cos (y/x) = k, where k = 1/C

This is the required solution of the given differential equation.

Question 8:

x * dy/dx – y + x * sin(y/x) = 0

The given differential equation is:

x * dy/dx – y + x * sin(y/x) = 0

=> x * dy/dx = y - x * sin(y/x)

=> dy/dx = {y - x * sin(y/x)}/x   …………1

Let F(x, y) = {y - x * sin(y/x)}/x

Now, F(kx, ky) = {ky - kx * sin(ky/kx)}/kx

=> F(kx, ky) = {y - x * sin(y/x)}/x

=> F(kx, ky) = k0 * F(x, y)

This shows that equation 1 is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Differentiating both sides with respect to x, we get:

dy/dx = v + x * dv/dx

Substituting the values of y and dy/dx in equation 1, we get:

v + x * dv/dx = {vx - x * sin v}/x

=> v + x * dv/dx = v - sin v

=> x * dv/dx = -sin v

=> -dv/sin v = dx/x

=> -cosec v dv = dx/x

Integrating both sides, we get:

=> log |cosec v – cot v| = -log x + Log C

=> log|(cosec v – cot v) = log (C/x)

=> cosec v – cot v = C/x

=> cosec (y/x) – cot (y/x) = C/x

=> 1/sin (y/x) – cos (y/x)/ sin (y/x) = C/x

=> x[1 – cos (y/x)] = C sin (y/x)

This is the required solution of the given differential equation.

Question 9:

y * dx + x * log(y/x)dy – 2x dy = 0

The given differential equation is:

y * dx + x * log(y/x)dy – 2x dy = 0

=> y * dx + [2x - x * log(y/x)]dy

=> dy/dx = y/[2x - x * log(y/x)]   …………1

Let F(x, y) = ky/[2kx - kx * log(ky/kx)]

Now, F(kx, ky) = y/[2x - x * log(y/x)]

=> F(kx, ky) = y/[2x - x * log(y/x)]

=> F(kx, ky) = k0 * F(x, y)

This shows that equation 1 is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Differentiating both sides with respect to x, we get:

dy/dx = v + x * dv/dx

Substituting the values of y and dy/dx in equation 1, we get:

v + x * dv/dx = vx/[2x - x * log v]

=> v + x * dv/dx = v/[2 - log v]

=> x * dv/dx = v/[2 - log v] - v

=> x * dv/dx = [v – 2v + v log v]/[2 - log v]

=> x * dv/dx = [v log v - v]/[2 - log v]

=> [2 – log v]/[v log v - v]dv = dx/x

=> [1 + 1 – log v]/[v(log v - 1)]dv = dx/x

=> [1/[v(log v - 1) – 1/v]dv = dx/x

Integrating both sides, we get:

=> ꭍ [1/[v(log v - 1) dv – ꭍ 1/v dv = ꭍ dx/x

=> ꭍ [1/[v(log v - 1) dv – log v = log x + log C  ………..2

Let log v -1 = t

=> d(log v -1)/dt = dt/dv

=> 1/v = dt/dv

=> dv/v = dt

Therefore, equation 2 becomes

=> ꭍ dt/t – log v = log x + log C

=> log t – log (y/x) = log (C/x)

=> log [log (y/x) - 1] – log (y/x) = log (C/x)

=> log [{log (y/x) – 1}/(y/x)] = log (C/x)

=> x{log (y/x) – 1}/y = C/x

=> log (y/x) – 1 = Cy

This is the required solution of the given differential equation.

Question 10:

(1 + ex/y)dx + ex/y (1 – x/y)dy = 0

The given differential equation is:

(1 + ex/y)dx + ex/y (1 – x/y)dy = 0

=> (1 + ex/y)dx = -ex/y (1 – x/y)dy

=> dx/dy = -ex/y (1 – x/y)/(1 + ex/y)   …………1

Let F(x, y) = -ex/y (1 – x/y)/(1 + ex/y)

Now, F(kx, ky) = -ekx/ky(1 – kx/ky)/(1 + ekx/ky)

=> F(kx, ky) = -ex/y (1 – x/y)/(1 + ex/y)

=> F(kx, ky) = k0 * F(x, y)

This shows that equation 1 is a homogeneous equation.

To solve it, we make the substitution as:

x = vy

Differentiating both sides with respect to y, we get:

dx/dy = v + y * dv/dy

Substituting the values of y and dx/dy in equation 1, we get:

v + y * dv/dy = -ev(1 – v)/(1 + ev)

=> y * dv/dy = -ev(1 – v)/(1 + ev) - v

=> y * dv/dy = (-ev + vev – v - vev)/(1 + ev)

=> y * dv/dy = -(v + ev)/(1 + ev)

=> [(1 + ev)/(v + ev)]dv = -dy/y

Integrating both sides, we get:

=> log (v + ev) = - log y + log C = log (C/y)

=> v + ev = C/y

=> x/y + ex/y = C/y

=> x + y * ex/y = C

This is the required solution of the given differential equation.

Question 11:

(x + y)dy + (x - y)dx  = 0; y = 1 when x = 1

The given differential equation is:

(x + y)dy + (x - y)dx  = 0

=> (x + y)dy = -(x - y)dx

=> dy/dx = -(x - y)/(x + y)   …………1

Let F(x, y) = -(x - y)/(x + y)

Now, F(kx, ky) = -(kx - ky)/(kx + ky)

=> F(kx, ky) = -(x - y)/(x + y)

=> F(kx, ky) = k0 * F(x, y)

This shows that equation 1 is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Differentiating both sides with respect to x, we get:

dy/dx = v + x * dv/dx

Substituting the values of y and dy/dx in equation 1, we get:

v + x * dv/dx = -(x - vx)/(x + vx)

=> v + x * dv/dx = (v - 1)/(v + 1)

=> x * dv/dx = (v - 1)/(v + 1) - v

=> x * dv/dx = [v – 1 – v2  - v]/ (v + 1)

=> x * dv/dx = -(1 + v2)/(v + 1)

=> [(v + 1)/(1 + v2)]dv = -dx/x

=> [v/(1 + v2) + 1/(1 + v2)]dv = -dx/x

Integrating both sides, we get:

=> (1/2) * log (1 + v2) + tan-1 v = -log x + k

=> log (1 + v2) + 2tan-1 v = -2log x + 2k

=> log [(1 + v2) * x2] + 2tan-1 v = 2k

=> log [(1 + y2/ x2) * x2] + 2tan-1 (y/x) = 2k

=> log (x2 + y2) + 2tan-1 (y/x) = 2k  ……….2

Now, y = 1, when x = 1

=> log 2 + 2tan-1 1 = 2k

=> log 2 + 2 * π/4 = 2k

=> log 2 + π/2 = 2k

Put the value of 2k in equation 2, we get

=> log (x2 + y2) + 2tan-1 (y/x) = log 2 + π/2

This is the required solution of the given differential equation.

Question 12:

x2 dy + (xy + y2)dx  = 0; y = 1, when x = 1

The given differential equation is:

x2 dy + (xy + y2)dx  = 0

=> x2 dy = -(xy + y2)dx

=> dy/dx = -(xy + y2)/x2   …………1

Let F(x, y) = -(xy + y2)/x2

Now, F(kx, ky) = -{kx * ky + (kx)2}/(kx)2

=> F(kx, ky) = -(xy + y2)/x2

=> F(kx, ky) = k0 * F(x, y)

This shows that equation 1 is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Differentiating both sides with respect to x, we get:

dy/dx = v + x * dv/dx

Substituting the values of y and dy/dx in equation 1, we get:

v + x * dv/dx = -{x * vx + (vx)2}/x2

=> v + x * dv/dx = -v – v2

=> x * dv/dx = -v – v2 - v

=> x * dv/dx = - v2 – 2v

=> x * dv/dx = -v(v + 2)

=> dv/{v(v + 2)} = -dx/x

=> [(1/2) * {v + 2 - v}/{v(v + 2)}]dv = -dx/x

=> (1/2) * [1/v - 1/(v + 2)]dv = -dx/x

Integrating both sides, we get:

=> (1/2) * [log v – log(v + 2) = -log x + log C

=> (1/2) * log [v/(v + 2)] = log (C/x)

=> log [v/(v + 2)] = 2 log (C/x)

=> log [v/(v + 2)] = log (C/x)2

=> v/(v + 2) = (C/x)2

=> (y/x)/(y/x + 2) = (C/x)2

=> y/(y + 2x) = C2/x2

=> x2 y/(y + 2x) = C2   …………2

Now, t = 1 at x = 1

=> 1/(1 + 2) = C2

=> C2 = 1/3

Put the value of C2 in equation 2, we get

=> x2 y/(y + 2x) = 1/3

=> y + 2x = 3x2 y

This is the required solution of the given differential equation.

Question 13:

[x sin2 (x/y) - y]dx + x dy  = 0; y = π/4, when x = 1

The given differential equation is:

[x sin2 (x/y) - y]dx + x dy

=> dy/dx = -[x sin2 (x/y) - y]/x   …………1

Let F(x, y) = -[x sin2 (x/y) - y]/x

Now, F(kx, ky) = -[kx sin2 (kx/ky) - ky]/kx

=> F(kx, ky) = -[x sin2 (x/y) - y)]/x

=> F(kx, ky) = k0 * F(x, y)

This shows that equation 1 is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Differentiating both sides with respect to x, we get:

dy/dx = v + x * dv/dx

Substituting the values of y and dy/dx in equation 1, we get:

v + x * dv/dx = -[x sin2 v - vx]/x

=> v + x * dv/dx = -[sin2 v - v]

=> x * dv/dx = v – sin2 v - v

=> x * dv/dx = - sin2 v

=> dv/ sin2 v = -dx/x

=> cosec2 v dv = -dx/x

Integrating both sides, we get:

=> -cot v = -log |x| - log C

=> cot v = log |x| + log C

=> cot (y/x) = log |Cx|   ……….2

Now, y = π/4, when x = 1

=> cot (π/4) = log |C|

=> 1 = log C

=> C = e1 = e

Put the value of C in equation 2, we get

cot (y/x) = log|ex|

This is the required solution of the given differential equation.

Question 14:

dy/dx – y/x + cosec (y/x)  = 0; y = 0, when x = 1

The given differential equation is:

dy/dx – y/x + cosec (y/x)  = 0

=> dy/dx = y/x - cosec (y/x)     …………1

Let F(x, y) = y/x - cosec (y/x)

Now, F(kx, ky) = ky/kx - cosec (ky/kx)

=> F(kx, ky) = y/x - cosec (y/x)

=> F(kx, ky) = k0 * F(x, y)

This shows that equation 1 is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Differentiating both sides with respect to x, we get:

dy/dx = v + x * dv/dx

Substituting the values of y and dy/dx in equation 1, we get:

v + x * dv/dx = v - cosec v

=> x * dv/dx = -cosec v

=> dv/cosec v = -dx/x

=> -sin v dv = dx/x

Integrating both sides, we get:

=> cos v = log |x| + log C

=> cos v = log |Cx|

=> cos (y/x) = log |Cx|   ……….2

Now, y = 0, when x = 1

=> cot (0) = log |C|

=> 1 = log C

=> C = e1 = e

Put the value of C in equation 2, we get

cos (y/x) = log|ex|

This is the required solution of the given differential equation.

Question 15:

2xy + y2 – 2x2 * dy/dx = 0; y = 2, when x = 1

The given differential equation is:

2xy + y2 – 2x2 * dy/dx = 0

=> 2x2 * dy/dx = 2xy + y2

=> dy/dx = (2xy + y2)/2x2    …………1

Let F(x, y) = (2xy + y2)/2x2

Now, F(kx, ky) = {2kx * ky + (ky)2}/2(kx)2

=> F(kx, ky) = (2xy + y2)/2x2

=> F(kx, ky) = k0 * F(x, y)

This shows that equation 1 is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Differentiating both sides with respect to x, we get:

dy/dx = v + x * dv/dx

Substituting the values of y and dy/dx in equation 1, we get:

v + x * dv/dx = {2x * vx + (vx)2}/2x2

=> v + x * dv/dx = (2v + v2)/2

=> v + x * dv/dx = v + v2/2

=> x * dv/dx = v2/2

=> (2/ v2)dv = dx/x

Integrating both sides, we get:

=> 2 * v-2 + 1/(-2 + 1) = log |x| + C

=> -2/v = log |x| + C

=> -2/(y/x) = log |x| + C

=> -2x/y = log |x| + C    ……….2

Now, y = 2, when x = 1

=> -1 = log |1| + C

=> C = -1

Put the value of C in equation 2, we get

=> -2x/y = log |x| - 1

=> 2x/y = 1 - log |x|

=> y = 2x/(1 – log |x|), where x ≠ 0, x ≠ e

This is the required solution of the given differential equation.

Question 16:

A homogeneous differential equation of the from dx/dy =  h(x/y) can be solved by making the substitution.                                                                                                                                               (A) y = vx                            (B) v = yx                             (C) x = vy                                (D) x = v

For solving the homogeneous equation of the form dx/dy =  h(x/y), we need to make the

substitution as x = vy.

Hence, the correct answer is C.

Question 17:

Which of the following is a homogeneous differential equation?

1. (4x + 6y + 5)dy – (3y + 2x + 4)dx = 0 B. xy dx – (x3 + y3)dy = 0
2. (x2 + 2y2)dx + 2xy dy = 0 D. y2 dx + (x2 - xy - y2)dy = 0

Function F(x, y) is said to be the homogenous function of degree n, if

F(λx, λy) = λn F(x, y) for any non-zero constant (λ).

Consider the equation given in alternative D:

y2 dx + (x2 - xy - y2)dy = 0

=> dy/dx = -y2/(x2 - xy - y2)

=> dy/dx = -y2/(y2 + xy - x2)

Let F(x, y) = -y2/(y2 + xy - x2)

Now, F(λx, λy) = -(λy)2/{(λy)2 + (λx)( λy) – (λx)2}

=> F(kx, ky) = - λ2y2/{λ2 * (y2 + xy2 - x2)}

=> F(kx, ky) = -y2/(y2 + xy2 - x2)

=> F(kx, ky) = k0 * F(x, y)

Hence, the differential equation given in alternative D is a homogenous equation.

Exercise 9.6

For each of the differential equations given in Exercises 1 to 12, find the general solution:

Question 1:

dy/dx + 2y = sin x

The given differential equation is:

dy/dx + 2y = sin x

This is in the form of dy/dx + py = Q, where p = 2 and Q = sin x

Now, I.F = eꭍp dx = eꭍ2 dx = e2x

The solution of the given differential equation is given by the relation,

y(I.F) = ꭍ(Q * I.F.) dx + C

=> y e2x = ꭍ sin x * e2x dx + C   ………..1

Let I = ꭍ sin x * e2x dx

=> I = sin x * ꭍ e2x dx - ꭍ[d(sin x/dx * ꭍ e2x dx]dx

=> I = sin x * e2x/2 - ꭍ [cos x * e2x/2]dx

=> I = sin x * e2x/2 – (1/2) * cos x * ꭍ e2x dx - ꭍ[d(cos x/dx * ꭍ e2x dx]dx

=> I = sin x * e2x/2 – (1/2) * [cos x * e2x/2 - ꭍ {(-sin x) * e2x/2]dx]

=> I = (sin x * e2x)/2 – (cos x * e2x)/4 – I/4

=> I + I/4 = e2x (2 sin x – cos x)/4

=> 5I/4 = e2x (2 sin x – cos x)/4

=> I = e2x (2 sin x – cos x)/5

Therefore, equation 1 becomes:

=> y e2x = e2x (2 sin x – cos x)/5 + C

=> y = (2 sin x – cos x)/5 + C e-2x

This is the required general solution of the given differential equation.

Question 2:

dy/dx + 3y = e-2x

The given differential equation is:

dy/dx + 3y = e-2x

This is in the form of dy/dx + py = Q, where p = 3 and Q = e-2x

Now, I.F = eꭍp dx = eꭍ3 dx = e3x

The solution of the given differential equation is given by the relation,

y(I.F) = ꭍ(Q * I.F.) dx + C

=> ye3x = ꭍ (e-2x * e3x) dx + C

=> ye3x = ꭍ ex dx + C

=> ye3x = ex + C

=> y = e-2x + Ce-3x

This is the required general solution of the given differential equation.

Question 3:

dy/dx + y/x = x2

The given differential equation is:

dy/dx + y/x = x2

This is in the form of dy/dx + py = Q, where p = 1/x and Q = x2

Now, I.F = eꭍp dx = eꭍ dx/x = elog x = x

The solution of the given differential equation is given by the relation,

y(I.F) = ꭍ(Q * I.F.) dx + C

=> y * x = ꭍ (x2 * x )dx + C

=> xy = ꭍ x3 dx + C

=> xy = x4/4 + C

This is the required general solution of the given differential equation.

Question 4:

dy/dx + y * sec x = tan x (0 ≤ x ≤ π/2)

The given differential equation is:

dy/dx + y * sec x = tan x

This is in the form of dy/dx + py = Q, where p = sec x and Q = tan x

Now, I.F = eꭍp dx = eꭍ sec x dx = elog (sec x + tan x) = sec x + tan x

The solution of the given differential equation is given by the relation,

y(I.F) = ꭍ(Q * I.F.) dx + C

=> y(sec x + tan x) = ꭍ tan x(sec x + tan x)dx + C

=> y(sec x + tan x) = ꭍ tan x * sec x dx + ꭍ tan2 x dx + C

=> y(sec x + tan x) = sec x + ꭍ(sec2 x – 1) dx + C

=> y(sec x + tan x) = sec x + tan x – x + C

This is the required general solution of the given differential equation.

Question 5:

0π/2 cos 2x dx

Let I = ꭍ0π/2 cos 2x dx

=> I = [sin 2x /2 0] π/2

=> I = (1/2) * [sin π – sin 0]

=> I = 0

Hence, ꭍ0π/2 cos 2x dx = 0

Question 6:

x * dy/dx + 2y = x2 log x

The given differential equation is:

dy/dx + 2y/x = x2 log x

This is in the form of dy/dx + py = Q, where p = 2/x and Q = x log x

Now, I.F = eꭍp dx = eꭍ 2/x dx = e2 log x = x2

The solution of the given differential equation is given by the relation,

y(I.F) = ꭍ(Q * I.F.) dx + C

=> y * x2 = ꭍ (x log x * x2)dx + C

=> x2y = ꭍ x3 log x dx + C

=> x2y = log x * ꭍ x3 dx - ꭍ[d(log x)/dx * ꭍ x3 dx]dx + C

=> x2y = log x * x4/4 - ꭍ[ 1/x *  x4/4]dx + C

=> x2y = log x * x4/4 – (1/4)ꭍ x3 + C

=> x2y = log x * x4/4 – (1/4) * x4/4 + C

=> x2y = x4 (4log x – 1)/16 + C

=> y = x2 (4log x – 1)/16 + Cx-2

This is the required general solution of the given differential equation.

Question 7:

x * log x dy/dx + y = 2/x log x

The given differential equation is:

x * log x dy/dx + y = 2/x log x

=> dy/dx + y/x log x = 2/x2

This is in the form of dy/dx + py = Q, where p = 1/x log x and Q = 2/x2

Now, I.F = eꭍp dx = eꭍ 1/x log x dx = elog(log x) = log x

The solution of the given differential equation is given by the relation,

y(I.F) = ꭍ(Q * I.F.) dx + C

=> y * log x = ꭍ(2/ x2 * log x)dx + C

=> y * log x = 2 * ꭍ log x/x2 dx + C

=> y * log x = 2[ log x * ꭍ 1/x2 dx  - ꭍ{d(log x)/dx * ꭍ1/x2 dx} dx] + C

=> y * log x = 2[ log x * (1-/x)  - ꭍ {(1/x) * (-1/x)} dx] + C

=> y * log x = 2[-log x/x + ꭍ 1/x2 dx] + C

=> y * log x = 2[-log x/x – 1/x] + C

=> y * log x = -2[1 + log x/x]/x + C

This is the required general solution of the given differential equation.

Question 8:

(1 + x2)dy + 2xy dx = cot x dx (x ≠ 0)

The given differential equation is:

(1 + x2)dy + 2xy dx = cot x dx

=> dy/dx + 2xy/(1 + x2) = cot x/(1 + x2)

This is in the form of dy/dx + py = Q, where p = 2x/(1 + x2) and Q = cot x/(1 + x2)

Now, I.F = eꭍp dx = eꭍ 2x/(1 + x2) dx = elog((1 + x2)) = 1 + x2

The solution of the given differential equation is given by the relation,

y(I.F) = ꭍ(Q * I.F.) dx + C

=> y * (1 + x2) = ꭍ[ cot x/(1 + x2) * (1 + x2)]dx + C

=> y(1 + x2) = ꭍ cot x dx + C

=> y(1 + x2) = log |sin x| + C

This is the required general solution of the given differential equation.

Question 9:

x * dy/dx + y – x + xy cot x = 0 (x ≠ 0)

The given differential equation is:

x * dy/dx + y – x + xy cot x = 0

=> x * dy/dx + y(1 + x cot x) = x

=> dy/dx + y(1/x + cot x) = 1

This is in the form of dy/dx + py = Q, where p = 1/x + cot x and Q = 1

Now, I.F = eꭍp dx = eꭍ (1/x + cot x)dx = elog x + log(sin x) = elog(x sin x) = x sin x

The solution of the given differential equation is given by the relation,

y(I.F) = ꭍ(Q * I.F.) dx + C

=> y * (x sin x) = ꭍ(1 * x sin x)dx + C

=> y(x sin x) = x ꭍ sin x dx  - ꭍ[d(x)/dx * ꭍsin x dx]dx + C

=> y(x sin x) = x(-cos x)  - ꭍ{1 * ꭍ(-cos x) dx} + C

=> y(x sin x) = -x cos x  + sin x + C

=> y(x sin x) = -x cos x/(x sin x)  + sin x/(x sin x) + C/(x sin x)

=> y(x sin x) = -cot x  + 1/x + C/(x sin x)

This is the required general solution of the given differential equation.

Question 10:

(x + y)dy/dx = 1

The given differential equation is:

(x + y)dy/dx = 1

=> dy/dx = 1/(x + y)

=> dx/dy = x + y

=> dx/dy - x = y

This is in the form of dx/dy + px = Q, where p = -1 and Q = y

Now, I.F = eꭍp dy = eꭍ-dy = e-y

The solution of the given differential equation is given by the relation,

x(I.F) = ꭍ(Q * I.F.) dy + C

=> x e-y = ꭍ (y * e-y)dy + C

=> xe-y = y ꭍ e-y dy – [ ꭍ d(y)/dy * ꭍe-y dy]dy + C

=> xe-y = -ye-y – ꭍ (-e-y)dy + C

=> xe-y = -ye-y + ꭍ e-y dy + C

=> xe-y = -ye-y - e-y + C

=> x = -y - 1 + Cey

=> x  + y + 1 = Cey

This is the required general solution of the given differential equation.

Question 11:

y dx + (x – y2)dy = 1

The given differential equation is:

y dx + (x – y2)dy = 0

=> y dx = (y2 – x)dy

=> dx/dy = (y2 – x)/y

=> dx/dy = y – x/y

=> dx/dy + x/y = y

This is in the form of dx/dy + px = Q, where p = 1/y and Q = y

Now, I.F = eꭍp dy = eꭍ1/y dy = elog y = y

The solution of the given differential equation is given by the relation,

x(I.F) = ꭍ(Q * I.F.) dy + C

=> xy = ꭍ (y * y)dy + C

=> xy = ꭍ y2 dy + C

=> xy = y3/3 + C

=> x = y2/3 + C/y

This is the required general solution of the given differential equation.

Question 12:

(x + 3y2) * dy/dx = y (y > 0)

The given differential equation is:

(x + 3y2) * dy/dx = y

=> dy/dx = y/(x + 3y2)

=> dx/dy = (x + 3y2)/y

=> dx/dy = x/y + 3y

=> dx/dy - x/y = 3y

This is in the form of dx/dy + px = Q, where p = -1/y and Q = 3y

Now, I.F = eꭍp dy = eꭍ-dy/y = e-log y = elog (1/y) = 1/y

The solution of the given differential equation is given by the relation,

x(I.F) = ꭍ(Q * I.F.) dy + C

=> x * 1/y = ꭍ (3y * 1/y)dy + C

=> x/y = 3ꭍ dy + C

=> x/y = 3y + C

=> x = 3y2 + Cy

This is the required general solution of the given differential equation.

Question 13:

dy/dx + 2y tan x = sin x; y = 0 when x = π/3

The given differential equation is:

dy/dx + 2y tan x = sin x

This is in the form of dy/dx + py = Q, where p = 2 tan x and Q = sin x

Now, I.F = eꭍp dy = eꭍ2 tan x dx = e 2log |sec x| = elog (sec2 x) = sec2 x

The solution of the given differential equation is given by the relation,

y(I.F) = ꭍ(Q * I.F.) dx + C

=> y * sec2 x = ꭍ (sin x * sec2 x)dx + C

=> y * sec2 x = ꭍ (sec x * tan x)dx + C

=> y * sec2 x = sec x + C  …………1

Now, y = 0 when x = π/3

=> 0 * sec2 π/3 = sec π/3 + C

=> 0 = 2 + C

=> C = -2

Put the value of C in equation 1, we get

=> y * sec2 x = sec x – 2

=> y = cos x – 2 cos2 x

This is the required general solution of the given differential equation.

Question 14:

(1 + x2)dy/dx + 2xy = 1/(1 + x2); y = 0 when x = 1

The given differential equation is:

(1 + x2)dy/dx + 2xy = 1/(1 + x2)

=> dy/dx + 2xy/(1 + x2) = 1/(1 + x2)2

This is in the form of dy/dx + py = Q, where p = 2 tan x and Q = sin x

Now, I.F = eꭍp dy = eꭍ 2xy/(1 + x2) dx = e log (1 + x2) = (1 + x2)

The solution of the given differential equation is given by the relation,

y(I.F) = ꭍ(Q * I.F.) dx + C

=> y(1 + x2) = ꭍ [1/(1 + x2) * (1 + x2)2]dx + C

=> y/(1 + x2) = ꭍ 1/(1 + x2) dx + C

=> y/(1 + x2) = tan-1 x + C  ………..1

Now, y = 0 when x = 1

=> 0 = tan-1 1 + C

=> C = -π/4

Put the value of C in equation 1, we get

=> y/(1 + x2) = tan-1 x - π/4

This is the required general solution of the given differential equation.

Question 15:

dy/dx - 3y cot x = sin 2x; y = 2 when x = π/2

The given differential equation is:

dy/dx - 3y cot x = sin 2x

This is in the form of dy/dx + py = Q, where p = -3 cot x and Q = sin 2x

Now, I.F = eꭍp dy = eꭍ -3 cot x dx = e3 log|sin x| = elog|1/sin3 x| = 1/sin3 x

The solution of the given differential equation is given by the relation,

y(I.F) = ꭍ(Q * I.F.) dx + C

=> y/sin3 x = ꭍ[sin 2x * 1/sin3 x]dx + C

=> y cosec3 x = 2ꭍ[cot x * cosec x]dx + C

=> y cosec3 x = 2 cosec x + C

=> y = -2/cosec2 x + C/cosec3 x

=> y = -2 sin2 x + C sin3 x  ………….1

Now, y = 2 when x = π/2

=> 2 = -2 + C

=> C = 4

Put the value of C in equation 1, we get

=> y = -2 sin2 x + 4 sin3 x

=> y = 4 sin3 x - 2 sin2 x

This is the required general solution of the given differential equation.

Question 16:

Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

Let F (x, y) be the curve passing through the origin.

At point (x, y), the slope of the curve will be dy/dx

According to the given information:

dy/dx = x + y

=> dy/dx - y = x

This is a linear differential equation of the form:

dy/dx + py = Q, where p = -1 and Q = x

Now, I.F = eꭍp dy = eꭍ (-1) dx = e-x

The solution of the given differential equation is given by the relation,

y(I.F) = ꭍ(Q * I.F.) dx + C

=> y(I.F) = ꭍ(x * e-x) dx + C

=> ye-x = x ꭍ e-x dx - ꭍ[d(x)/dx * ꭍ e-x dx]dx + C

=> ye-x = -xe-x - ꭍ(-e-x) dx + C

=> ye-x = -xe-x + ꭍ e-x dx + C

=> ye-x = -xe-x - e-x + C

=> ye-x = -e-x (x + 1) + C

=> y = -(x + 1) + Cex

=> x + y + 1 = Cex  …………..1

The curve passes through the origin.

=> 1 = Ce0

=> C = 1

Put the value of C in equation 2, we get

x + y + 1 = ex

Hence, the required equation of curve passing through the origin is x + y + 1 = ex

Question 17:

Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Let F (x, y) be the curve and let (x, y) be a point on the curve. The slope of the tangent to the

curve at (x, y) is dy/dx

According to the given information:

dy/dx + 5 = x + y

=> dy/dx - y = x – 5

This is a linear differential equation of the form:

dy/dx + py = Q, where p = -1 and Q = x - 5

Now, I.F = eꭍp dy = eꭍ (-1) dx = e-x

The solution of the given differential equation is given by the relation,

y(I.F) = ꭍ(Q * I.F.) dx + C

=> y(I.F) = ꭍ(x – 5) * e-x dx + C

=> ye-x = (x – 5)ꭍ e-x dx - ꭍ[d(x – 5)/dx * ꭍ e-x dx]dx + C

=> ye-x = -(x – 5)e-x - ꭍ(-e-x) dx + C

=> ye-x = (5 - x)e-x + ꭍ e-x dx + C

=> ye-x = (5 - x)e-x - e-x + C

=> ye-x = (4 - x)e-x + C

=> y = 4 - x + Cex

=> x + y - 4 = Cex …………….1

The curve passes through the point (0, 2).

=> 0 + 2 - 4 = Ce0

=> C = -2

Put the value of C in equation 2, we get

x + y - 4 = -2ex

=> y = 4 – x - 2ex

This is the required equation of the curve.

Question 18:

The integrating factor of the differential equation is x * dy/dx – y = 2x2                                                                        A. e–x                        B. e–y                     C. 1/x                                    D. x

The given differential equation is:

x * dy/dx – y = 2x2

=> dy/dx – y/x = 2x

This is a linear differential equation of the form:

This is a linear differential equation of the form:

dy/dx + py = Q, where p = -1/x and Q = 2x

Now, I.F = eꭍp dy = eꭍ (-1/x) dx = e-log(x) = elog(x-1) = x-1 = 1/x

Hence, the correct answer is option C.

Question 19:

The integrating factor of the differential equation.

(1 - y2) * dy/dx + yx = ay  (-1< y < 1)

1. 1/(y2 - 1) B. 1/√(y2 - 1)         C. 1/(1 - y2)                            D. 1/√(1 - y2)

The given differential equation is:

(1 - y2) * dy/dx + yx = ay

=> dy/dx + yx/(1 - y2) = ay/(1 - y2)

This is a linear differential equation of the form:

dy/dx + py = Q, where p = 1/(1 - y2) and Q = ay/(1 - y2)

Now, I.F = e y/(1 - y2) dy = e-log(1 – y2)/2 = elog[1/√(1 – y2)} = 1/√(1 - y2)

Hence, the correct answer is option D.

Miscellaneous Exercise on Chapter 9

Question 1:

For each of the differential equations given below, indicate its order and degree (if defined).

(i) d2y/dx2 + 5x(dy/dx)2 – 6y = log x

(ii) (dy/dx)3 - 4(dy/dx)2 + 7y = sin x

(iii) d4y/dx4 – sin(d3y/dx3) = 0

(i) The differential equation is given as:

d2y/dx2 + 5x(dy/dx)2 – 6y = log x

=> d2y/dx2 + 5x(dy/dx)2 – 6y - log x = 0

The highest order derivative present in the differential equation is d2y/dx2.

Thus, its order is two. The highest power raised to is one. Hence, its degree is one.

(ii) The differential equation is given as:

(dy/dx)3 - 4(dy/dx)2 + 7y = sin x

=> (dy/dx)3 - 4(dy/dx)2 + 7y - sin x = 0

The highest order derivative present in the differential equation is dy/dx.

Thus, its order is one. The highest power raised to dy/dx is three. Hence, its degree is three.

(iii) The differential equation is given as:

d4y/dx4 – sin(d3y/dx3) = 0

The highest order derivative present in the differential equation is d4y/dx4.

Thus, its order is four.

However, the given differential equation is not a polynomial equation.

Hence, its degree is not defined.

Question 2:

For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(i) y = aex + be-x + x2                         :    x * d2y/dx2 + 2(dy/dx) – xy + x2 - 2 = 0

(ii) y = ex(a cos x + b sin x)            :      d2y/dx2 - 2(dy/dx) + 2y = 0

(iii) y = x sin 3x                                  :    d2y/dx2 + 9y – 6 cos 3x = 0

(iv) x2 = 2y2 log y                               :     (x2 + y2) * dy/dx – xy = 0

(i) Given, y = aex + be-x + x2

Differentiating both sides with respect to x, we get:

dy/dx = aex - be-x + 2x

Again, differentiating both sides with respect to x, we get:

d2y/dx2 = aex + be-x + 2

Now, on substituting the values of dy/dx and d2y/dx2 in the differential equation, we get:

LHS:

x * d2y/dx2 + 2(dy/dx) – xy + x2 – 2

= x * (aex + be-x + 2) + 2(aex - be-x + 2x) – x(aex + be-x + x2) + x2 – 2

= axex + bxe-x + 2x + 2aex - 2be-x + 4x – axex - bxe-x - x3 + x2 – 2

= axex + bxe-x - x3 + x2 + 6x - 2

≠ R.H.S.

Hence, the given function is not a solution of the corresponding differential equation.

(ii) Given, y = aex(a cos x + b sin x)

=> y = aex cos x + bex sin x

Differentiating both sides with respect to x, we get:

dy/dx = a d(ex cos x)/dx + b d(ex sin x)/dx

=> dy/dx = a(ex cos x - ex sin x) + b (ex sin x + ex cos x)

=> dy/dx = (a + b)ex cos x + (b - a)ex sin x

Again, differentiating both sides with respect to x, we get:

=> d2y/dx2 = (a + b) d(ex cos x)/dx + (b - a) d(ex sin x)/dx

=> d2y/dx2 = (a + b) [ex cos x - ex sin x] + (b - a)[ex sin x + ex cos x]

=> d2y/dx2 = ex[(a + b)(cos x - sin x) + (b - a)(sin x + cos x)]

=> d2y/dx2 = 2ex(b cos x – a sin x)

Now, on substituting the values of d2y/dx2 and dy/dx in the LHS of the given differential

equation, we get:

d2y/dx2 - 2(dy/dx) + 2y

= 2ex(b cos x – a sin x) – 2[(a + b)ex cos x + (b - a)ex sin x] + 2aex(a cos x + b sin x)

= ex[(2b cos x – 2a sin x) - (2a cos x + 2b cos x) - (2b sin x – 2a sin x) + (2a cos x + 2b sin x)]

= ex[(2b – 2a – 2b + 2a)cos x + (-2a – 2b + 2a + 2b)sin x]

= 0

Hence, the given function is a solution of the corresponding differential equation.

(iii) Given y = x sin 3x

Differentiating both sides with respect to x, we get:

dy/dx = d(x sin 3x)/dx

=> dy/dx = sin 3x + 3x cos 3x

Again, differentiating both sides with respect to x, we get:

=> d2y/dx2 = d(sin 3x)/dx + d(3x cos 3x)/dx

=> d2y/dx2 = 3 cos 3x + 3[cos 3x + x(-sin 3x) * 3]

=> d2y/dx2 = 6 cos 3x – 9x sin 3x

Put the value of d2y/dx2 in the L.H.S. of the given differential equation, we get:

d2y/dx2 + 9y – 6 cos 3x

= 6 cos 3x – 9x sin 3x + 9y – 6 cos 3x

= 0

Hence, the given function is a solution of the corresponding differential equation.

(iv) Given, x2 = 2y2 log y

Differentiating both sides with respect to x, we get:

2x = 2 * d(y2 log y)/dx

=> x = [2y * log y * dy/dx + y2 * 1/y * dy/dx]

=> x = [2y log y + y] * dy/dx

=> dy/dx = x/(2y log y + y)

=> dy/dx = x/[y(1 + 2 log y)]

Put the value of dy/dx in the L.H.S. of the given differential equation, we get:

(x2 + y2) * dy/dx – xy

= (2y2 log y + y2) * x/[y(1 + 2 log y)] – xy

= y2 (1 + 2 log y) * x/[y(1 + 2 log y)] – xy

= xy – xy

= 0

Hence, the given function is a solution of the corresponding differential equation.

Question 3:

Form the differential equation representing the family of curves given by                                            (x - a)2 + 2y2 = a2 where a is an arbitrary constant.

Given, (x - a)2 + 2y2 = a2

=> x2 - a2 – 2ax + 2y2 = a2

=> 2y2 = 2ax - x2  ………1

Differentiating with respect to x, we get:

2y * dy/dx = (2a – 2x)/2

=> dy/dx = (a – x)/2y

=> dy/dx = (2ax – 2x2)/4xy   ………2

From equation 1, we get:

2ax = 2y2 + x2

On substituting this value in equation 2, we get:

=> dy/dx = (2y2 + x2 – 2x2)/4xy

=> dy/dx = (2y2 - x2)/4xy

Hence, the differential equation of the family of curves is given as dy/dx = (2y2 - x2)/4xy

Question 4:

Prove that (x2 - y2) = c(x2 - y2)2 is the general solution of differential equation   (x3 – 3xy2)dx = (y3 – 3x2y)dy, where c is a parameter.

Given, (x3 – 3xy2)dx = (y3 – 3x2y)dy

=> dy/dx = (x3 – 3xy2)/(y3 – 3x2y)   ………..1

This is a homogeneous equation. To simplify it, we need to make the substitution as:

y = vx

=> dy/dx = v + x * dv/dx

Put the value of y and dy/dx in equation 1, we get

v + x * dv/dx = {x3 – 3x(vx)2}/{(vx)3 – 3x2y(vx)}

=> v + x * dv/dx = (1 – 3v2)/(v3 – 3v)

=> x * dv/dx = (1 – 3v2)/(v3 – 3v) – v

=> x * dv/dx = {(1 – 3v2) – v(v3 – 3v)}/(v3 – 3v)

=> x * dv/dx = (1 – v4)/(v3 – 3v)

=> {(v3 – 3v)/(1 – v4)}dv = dx/x

Integrating both sides, we get:

=> ꭍ{(v3 – 3v)/(1 – v4)}dv = ꭍ dx/x

=> ꭍ[v3/(1 – v4)]dv - 3ꭍ[v/(1 – v4)dv = log x + log C   ………….2

Let ꭍ{(v3 – 3v)/(1 – v4)}dv = A – 3B where A = ꭍ[v3/(1 – v4)]dv and B = ꭍ[v/(1 – v4)dv

Let 1 – v4 = t

=> d(1 – v4) = dt/dv

=> -4v3 = dt/dv

=> v3 dv = -dt/4

Now, A = (-1/4)ꭍ dt/t = (-1/4)log t = (-1/4)log (1 – v4)

And B = ꭍ[v/(1 – v4)dv = ꭍ[v/{1 – (v)2)2}dv

Let v2 = p

=> d(v2)/dv = dp/dv

=> 2v = dp/dv

=> v dv = dp/2

Now, B = (1/2) ꭍ dp/(1 – p2) = 1/(2 * 2) log|(1 + p)/(1 - p)| = (1/4) log|(1 + v2)/(1 - v2)|

Substituting the values of A and B in equation 2, we get:

(-1/4)log (1 – v4) - (3/4) log|(1 + v2)/(1 - v2)| = log x + log C

=> (-1/4)log|(1 – v4)  * (1 + v2)/(1 - v2)| = log Cx

=> (1 + v2)4/(1 - v2)2 = (Cx)-4

=> (1 + y2/x2)4/(1 - y2/x2)2 = (1/Cx)4

=> (x2 + y2)4/{x4(x2 - y2)2} = 1/C4x4

=> (x2 - y2)2 = C2(x2 + y2)4

=> x2 - y2 = C2(x2 + y2)2

Hence, the given result is proved.

Question 5:

Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

The equation of a circle in the first quadrant with centre (a, a) and radius (a) which touches the

coordinate axes is:

(x - a)2 + (y - a)2 = a2 ……….1

Differentiating equation 1 with respect to x, we get:

2(x - a) + 2(y - a) * dy/dx = 0

=> (x - a) + (y - a)y’ = 0

=> x – a + yy’ – ay’ = 0

=> x + yy’ – a(1 + y’) = 0

=> a = (x + yy’)/(1 + y’)

Substituting the value of a in equation 1, we get:

[x - (x + yy’)/(1 + y’)]2 + [y - (x + yy’)/(1 + y’)]2 = [(x + yy’)/(1 + y’)]2

=> [(x – y)y’)/(1 + y’)]2 + [(y - x)/(1 + y’)]2 = [(x + yy’)/(1 + y’)]2

=> (x – y)2 y’2 + (x - xy)2 = (x + yy’)2

=> (x – y)2 [1 + y’2] = (x + yy’)2

Hence, the required differential equation of the family of circles is (x – y)2 [1 + y’2] = (x + yy’)2

Question 6:

Find the general solution of the differential equation dy/dx + √{(1 – y2)/(1 – x2)} = 0

Given, dy/dx + √{(1 – y2)/(1 – x2)} = 0

=> dy/dx = -√{(1 – y2)/(1 – x2)}

=> dy/(1 – y2) = -dx/(1 – x2)

Integrating both sides, we get:

=> ꭍ dy/(1 – y2) = -ꭍ dx/(1 – x2)

=> sin-1 y = -sin-1 x + C

=> sin-1 y + sin-1 x = C

Question 7:

Show that the general solution of the differential equation dy/dx + (y2 + y + 1)/(x2 + x + 1) = 0

is given by (x + y + 1) = A (1 – x – y – 2xy), where A is parameter.

Given, dy/dx + (y2 + y + 1)/(x2 + x + 1) = 0

=> dy/dx = -(y2 + y + 1)/(x2 + x + 1)

=> dy/(y2 + y + 1) = -dx/(x2 + x + 1)

=> dy/(y2 + y + 1) + dx/(x2 + x + 1) = 0

Integrating both sides, we get:

=> ꭍ dy/(y2 + y + 1) + ꭍ dx/(x2 + x + 1) = C

=> ꭍ dy/{(y + 1/2)2 + (√3/2)2} + ꭍ dy/{(x + 1/2)2 + (√3/2)2} = C

=> (2/√3) tan-1 [(y + 1/2)/( √3/2)] + (2/√3) tan-1 [(x + 1/2)/( √3/2)] = C

=> tan-1 [(2y + 1)/√3] + tan-1 [(2x + 1)/√3] = √3C/2

=> tan-1 [{(2y + 1)/√3 + (2x + 1)/√3}/{1 - (2y + 1)/√3 * (2x + 1)/√3}] = √3C/2

=> tan-1 [{(2x + 2y + 2)/√3}/{1 – (4xy + 2x + 2y + 1)/3}] = √3C/2

=> tan-1 [2√3(2x + 2y + 2)/{2(1 - x - y – 2xy)}] = √3C/2

=> √3(2x + 2y + 2)/{2(1 - x - y – 2xy)}] = tan (√3C/2)

=> √3(2x + 2y + 2)/{2(1 - x - y – 2xy)}] = B, where B = tan (√3C/2)

=> x + y + 1 = 2B(1 - x - y – 2xy)/√3

=> x + y + 1 = A(1 - x - y – 2xy), where A = 2B/√3

Hence, the given result is proved.

Question 8:

Find the equation of the curve passing through the point (0, π/4) whose differential equation is sin x cos y dx + cos x sin x dy = 0

The differential equation of the given curve is:

sin x cos y dx + cos x sin x dy = 0

=> (sin x cos y dx + cos x sin x dy)/(cos x cos y) = 0

=> tan x dx + tan y dy = 0

Integrating both sides, we get:

=> log(sec x) + log(sec y) = log C

=> log(sec x * sec y) = log C

=> sec x * sec y = C   …………1

The curve passes through point (0, π/4)

So, sec 0 * sec π/4 = C

=> C = 1 * √2

=> C = √2

Put value of C in equation 1, we get

sec x * sec y = √2

=> sec x * 1/cos y = √2

=> cos y = sec x /√2

Hence, the required equation of the curve is cos y = sec x /√2

Question 9:

Find the particular solution of the differential equation (1 + e2x)dy + (1 + y2)ex dx = 0 , given that y = 1 when x = 0.

Given, (1 + e2x)dy + (1 + y2)ex dx = 0

=> dy/(1 + y2) + ex  * dx/(1 + e2x) = 0

Integrating both sides, we get:

tan-1 y + ꭍ ex * dx/(1 + e2x) = 0

Let ex = t

=> e2x = t2

d(ex )/dx = dt/dx

=> ex = dt/dx

=> ex dx = dt

Substituting these values in equation 1, we get:

tan-1 y + ꭍ dt/(1 + t2) = C

=> tan-1 y + tan-1 t = C

=> tan-1 y + tan-1 ex = C   ………2

Now, y = 1 at x = 0

Therefore, equation 2 becomes:

tan-1 1 + tan-1 1 = C

=> C = π/4 + π/4

=> C = π/2

Put value of C in equation 2, we get

=> tan-1 y + tan-1 ex = π/2

This is the required particular solution of the given differential equation.

Question 10:

Solve the differential equation y * ex/y dx = (x * ex/y + y2)dy, y ≠ 0

Given, y * ex/y dx = (x * ex/y + y2)dy

=> y * ex/y * dx/dy = x * ex/y + y2

=> ex/y[y * dx/dy – x] = y2

=> ex/y[y * dx/dy – x]/y2 = 1  ………….1

Let ex/y = z

Differentiating it with respect to y, we get:

d(ex/y)/dy = dz/dy

=> ex/y [(y * dx/dy - x)/y2] = dz/dy   ……….2

From equation 1 and equation 2, we get:

dz/dy = 1

=> dz = dy

Integrating both sides, we get:

=> z = y + C

=> ex/y = y + C

Question 11:

Find a particular solution of the differential equation  (x - y)(dx + dy) = dx – dy, given that y = – 1, when x = 0 (Hint: put x – y = t)

Given, (x - y)(dx + dy) = dx – dy

=> (x – y + 1)dy = (1 – x + y)dy

=> dy/dx = (1 – x + y)/(x – y + 1)

=> dy/dx = (1 – x + y)/{1 + (x – y)}     ……….1

Let x – y = t

=> d(x - y)/dx = dt/dx

=> 1 – dy/dx = dt/dx

=> 1 – dt/dx = dy/dx

Substituting the values of x – y and dy/dx in equation 1, we get:

=> 1 – dt/dx = (1 - t)/(1 + t)

=> dt/dx = 1 - (1 - t)/(1 + t)

=> dt/dx = {(1 + t) - (1 - t)}/(1 + t)

=> dt/dx = 2t/(1 + t)

=> {(1 + t)/t} dt = 2 dx

=> (1 + 1/t) dt = 2 dx   ………2

Integrating both sides, we get:

t + log |t| = 2x + C

=> (x - y) + log |x - y| = 2x + C

=> log|x - y| = x + y + C

Now, y = –1 at x = 0

Therefore, equation (3) becomes:

log 1 = 0 – 1 + C

=> C = 1

Substituting C = 1 in equation 3 we get:

=> log|x - y| = x + y + 1

This is the required particular solution of the given differential equation.

Question 12:

Solve the differential equation [e-2√x/√x – y/√x](dx/dy) = 1, x ≠ 0

Given, [e-2√x/√x – y/√x](dy/dx) = 1

=> dy/dx = [e-2√x/√x – y/√x]

=> dy/dx + y/√x = e-2√x/√x

This equation is a linear differential equation of the form dy/dx + Py = Q

Where P = 1/√x and Q = e-2√x/√x

Now, IF = eꭍP dx = eꭍ dx/√x = e2√x

The general solution of the given differential equation is given by,

y(IF) = ꭍ (Q * IF) dx + C

=> ye2√x = ꭍ (e-2√x/√x * e2√x) dx + C

=> ye2√x = ꭍ dx/√x + C

=> ye2√x = 2√x + C

Question 13:

Find a particular solution of the differential equation

dy/dx + y cot x = 4x * cosec x, x ≠ 0, given that y = 0 when x = π/2

The given differential equation is:

dy/dx + y cot x = 4x * cosec x

This equation is a linear differential equation of the form dy/dx + Py = Q

Where P = cot x and Q = 4x * cosec x

Now, IF = eꭍ P dx = eꭍ cot x dx = elog|sin x| = sin x

The general solution of the given differential equation is given by,

y(IF) = ꭍ (Q * IF) dx + C

=> y* sin x = ꭍ (4x * cosec x * sin x) dx + C

=> y* sin x = 4 ꭍ x dx + C

=> y* sin x = 4 * x2/2 + C

=> y* sin x = 2x2 + C    ……….1

Now, y = 0 at x = π/2, equation 1 becomes

=> 0* sin x = 2(π/2)2 + C

=> C = -π2/2

Put value of C in equation 1, we get

=> y* sin x = 2x2 - π2/2

This is the required particular solution of the given differential equation.

Question 14:

Find a particular solution of the differential equation    (x + 1) * dy/dx = 2e-y - 1, given that y = 0 when x = 0

Given, (x + 1) * dy/dx = 2e-y – 1

=> dy/(2e-y – 1) = dx/(x + 1)

=> {ey/(2 – ey)}dy = dx/(x + 1)

Integrating both sides, we get:

=> ꭍ {ey/(2 – ey)}dy = log|x + 1| + log C   ………..1

Let 2 – ey = t

=> d(2 – ey)/dy = dt/dy

=> -ey = dt/dy

=> ey dy = -dt

Substituting this value in equation 1, we get:

=> ꭍ -dt/t = log|x + 1| + log C

=> -log|t| = log|C(x + 1)|

=> -log|2 – ey| = log|C(x + 1)|

=> 1/(2 – ey) = C(x + 1)

=> (2 – ey) = 1/{C(x + 1)}     ……….2

Now, at x = 0 and y = 0, equation 2 becomes:

=> (2 – e0) = 1/{C(0 + 1)}

=> 2 – 1 = 1/C

=> C = 1

Substituting C = 1 in equation 2, we get:

=> (2 – ey) = 1/(x + 1)

=> ey = 2 - 1/(x + 1)

=> ey = {2(x + 1) – 1}/(x + 1)

=> ey = (2x + 1)/(x + 1)

=> y = log|(2x + 1)/(x + 1)|, x ≠ 1

This is the required particular solution of the given differential equation.

Question 15:

The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time.

If the population of the village was 20000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

Let the population at any instant (t) be y.

It is given that the rate of increase of population is proportional to the number of inhabitants

at any instant.

So, dy/dt is proportional to y

=> dy/dt = ky, where k is a constant

=> dy/y = k dt

Integrating both sides, we get:

log y = kt + C    ………1

In the year 1999, t = 0 and y = 20000.

Therefore, we get:

log 20000 = C    …….2

In the year 2004, t = 5 and y = 25000

Therefore, we get:

log 25000 = k * 5 + C

=> log 25000 = k * 5 + log 20000

=> 5k = log 25000 - log 20000

=> 5k = log (25000/20000)

=> 5k = log (5/4)

=> k = (1/4) * log (5/4)    ………3

In the year 2009, t = 10 years.

Now, on substituting the values of t, k, and C in equation 1, we get:

log y = 10 * (1/5) * log (5/4) + log 20000

=> log y = 2 * log (5/4) + log 20000

=> log y = log (5/4)2 + log 20000

=> log y = log (25/16) + log 20000

=> log y = log (25/16 * 20000)

=> y = 25/16 * 20000

=> y = 31250

Hence, the population of the village in 2009 will be 31250.

Question 16:

The general solution of the differential equation is (y dx – x dy)/y = 0

A. xy = C                           B. x = Cy2                              C. y = Cx                           D. y = Cx2

The given differential equation is:

(y dx – x dy)/y = 0

=> (y dx – x dy)/xy = 0

=> dx/x – dy/y = 0

Integrating both sides, we get:

=> log|x| - log|y| = log k

=> log|x/y| = log k

=> k = xy

=> y = x/k

=> y = Cx, where C = 1/k

Hence, the correct answer is option C.

Question 17:

The general solution of a differential equation of the type is dx/dy + P1x = Q1 is

1. yeꭍ P1 dy = ꭍ(Q1 * eꭍ P1 dy )dy + C B. yeꭍ P1 dx = ꭍ(Q1 * eꭍ P1 dx )dx + C
2. xeꭍ P1 dy = ꭍ(Q1 * eꭍ P1 dy )dy + C D. xeꭍ P1 dx = ꭍ(Q1 * eꭍ P1 dx )dx + C

The integrating factor of the given differential equation dx/dy + P1x = Q1 is eꭍ P1 dx

The general solution of the differential equation is given by,

x(IF) = ꭍ(x *IF)dy + C

=> xeꭍ P1 dy = ꭍ(Q1 * eꭍ P1 dy )dy + C

Hence, the correct answer is option C.

Question 18:

The general solution of the differential equation ex dy + (yex + 2x)dy = 0 is

A. xey + x2 = C                 B. xey + y2 = C                  C. yex + x2 = C                 D. yey + x2 = C

The given differential equation is:

ex dy + (yex + 2x)dy = 0

=> ex * dy/dx + yex + 2x = 0

=> dy/dx + y = -2xe-x

This is a linear differential equation of the form dy/dx + Py = Q

Where P = 1 and Q = -2xe-x

Now, IF = eꭍ P dx = e ꭍ dx = ex

The general solution of the given differential equation is given by,

y(IF) = ꭍ (Q * IF) dx + C

=> yex = ꭍ (-2xe-x * ex) dx + C

=> yex = -ꭍ 2x dx + C

=> yex = -2x2/2 + C

=> yex = -x2 + C

=> yex + x2 = C

Hence, the correct answer is option C.