# Class 12 Maths NCERT Solutions for Chapter 9 Differential Equations Exercise 9.5

### Differential Equations Exercise 9.5 Solutions

**1. Show that the given differential equation is homogeneous and solve each of them(x ^{2} + xy) dy = (x^{2} + y^{2}) dx**

**Solution**

The given differential equation i.e. (x^{2} + xy) dy = (x^{2} + y^{2}) dx can be written as :

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 v and dy/dx in equation (1), we get :

Integrating both sides, we get :

-2 log (1 - v) - v = log x - log k

⇒ v = -2log (1 - v) - log x + log k

This is the required solution of the given differential equation.

**2. Show that the given differential equation is homogeneous and solve each of them y' = (x + y)/x **

**Solution**

The given differential equation is :

Thus, the given equation 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 :

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.

**3. Show that the given differential equation is homogeneous and solve each of them(x – y) dy – (x + y) dx = 0**

**Solution**

The given differential equation is :

(x - y)dy - (x + y)dx = 0

Thus, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as :

y = vx

Integrating both sides, we get :

This is the required solution of the given differential equation.

**4. Show that the given differential equation is homogeneous and solve each of them(x ^{2} – y^{2}) dx + 2xy dy = 0 **

**Solution**

The given differential equation is :

(x^{2} – y^{2}) dx + 2xy dy = 0

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as :

y = vx

Integrating both sides, we get :

log (1 + v)^{2} = -log x + log C = log (C/x)

⇒ 1 + v^{2} = C/x

⇒ x^{2} + y^{2} = Cx

This is the required solution of the given differential equation.

**5. Show that the given differential equation is homogeneous and solve each of themx ^{2} (dy/dx) = x^{2} - 2y^{2} + xy **

**Solution**

The given differential equation is :

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as :

y = vx

⇒ dy/dx = v + x(dv/dx)

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

Integrating both sides, we get:

This is the required solution for the given differential equation.

**6. Show that the given differential equation is homogeneous and solve each of themx dy -y dx = √(x ^{2} + y^{2})dx**

**Solution**

x dy -y dx = √(x^{2} + y^{2} )dx

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as :

y = vx

Substituting the values of v and dy/dx in equation (1), we get :

This is the required solution of the given differential equation.

**7. Show that the given differential equation is homogeneous and solve each of them{x cos (y/x) + y sin (y/x)} ydx = {y sin (y/x) - x cos (y/x)} x dy **

**Solution**

The given differential equation is :

= Î»^{0}. F(x, y)

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as :

y = vx

⇒ dy/dx = v + x = dv/dx

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

Integrating both sides, we get :

log (sec v) - log v = 2 log x + log C

This is the required solution of the given differential equation.

**8. Show that the given differential equation is homogeneous and solve x (dy/dx) - y + x sin (y/x) = 0 **

**Solution**

Therefore, the given differential equation is a homogeneous equation.

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

⇒ cosec v dv = -dx/x

Integrating both sides, we get :

log|cosecv - cot v| = -log x + log C = log (C/x)

This is the required solution of the given differential equation.

**9. Show that the given differential equation is homogeneous and solve**

y dx + x log (y/x)dy - 2xdy = 0

y dx + x log (y/x)dy - 2xdy = 0

**Solution**

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as :

y = vx

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

Integrating both sides, we get :

⇒ dv/v = dt

Therefore, equation (1) becomes :

⇒ ∫dt/t - log v = log x + log C

⇒ log t - log (y/x) = log (Cx)

This is the required solution of the given differential equation.

**10. Show that the given differential equation is homogeneous and solve**

(1 + e

(1 + e

^{x/y}) dx + e^{x/y}(1 - x/y)dy = 0**Solution**

Therefore, the given differential equation is a homogeneous equation.

x = vy

Substituting the values of x and dx/dy in equation (1), we get :

Integrating both sides, we get :

⇒ log (v + e

^{x}) = -log y + log C = log(C/y)

⇒ x + ye

^{x/y}= C

This is the required solution of the given differential equation.

**11. For the differential equations find the particular solution satisfying the given condition:**

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

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

**Solution**

⇒ (x + y)dy = -(x - y) dx

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as :

y = vx

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

Integrating both sides, we get :

Now, y = 1 at x = 1.

⇒ log 2 + 2 tan

^{-1}1 = 2k

⇒ log 2 + 2 × Ï€/4 = 2k

⇒ Ï€/2 + log 2 = 2k

log(x

^{2}+ y

^{2}) + 2 tan

^{-1}(y/x) = Ï€/2 + log 2

**12.For the differential equations find the particular solution satisfying the given condition:**

**x**

^{2}dy + (xy + y^{2}) dx = 0; y = 1 when x = 1**Solution**

^{2}dy + (xy + y

^{2})dx = 0

⇒ x

^{2}dy = -(xy + y

^{2})dx

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as :

y = vx

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

Integrating both sides, we get :

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

Now, y =1 at x = 1.

⇒ 1/(1+2) = C

^{2}

⇒ C

^{2}= 1/3

Substituting C

^{2}= 1/3 in equation (2), we get:

x

^{2}y/(y + 2x) = 1/3

⇒ y + 2x = 3x

^{2}y

This is the required solution of the given differential equation.

**13. For the differential equations find the particular solution satisfying the given condition:**

[x sin

[x sin

^{2}(y/x - y)] dx + xdy = 0; y = Ï€/4 when x = 1**Solution**

Therefore, the given differential equation is a homogeneous equation.

To solve this differential equation, we make the substitution as :

y = vx

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

Integrating both sides, we get :

- cot v = -log |x| - C

⇒ cot v = log |x| + C

⇒ cot(y/x) = log |x| + log C

⇒ cot (y/x) = log |Cx|

**...(2)**

Now, y = Ï€/4 at x = 1.

⇒ cot (Ï€/4) = log |C|

⇒ 1 = log C

⇒ C = e

^{1}= e

Substituting C = e in equation (2), we get :

cot (y/x) = log |ex|

This is the required solution of the given differential equation .

**14. For the differential equations find the particular solution satisfying the given condition:**

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

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

**Solution**

y = vx

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

Integrating both sides, we get :

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

⇒ cos (y/x) = log |Cx|

**...(2)**

This is the required solution of the given differential equation.

Now, y = 0 at x = 1.

⇒ cos (0) = log C

⇒ 1 = log C

⇒ C = e

^{1}= e

Substituting C = e in equation (2), we get :

cos(y/x) = log |(ex)|

This is the required solution of the given differential equation.

**15. For the differential equations find the particular solution satisfying the given condition:**

2xy + y

2xy + y

^{2}- 2x^{2}(dy/dx) = 0; y = 2 when x = 1**Solution**

Therefore, the given differential equation is a homogeneous equation.

y = vx

Substituting the value of y and dy/dx in equation (1), we get :

Now, y = 2 at x = 1.

⇒ C = -1

Substituting C = -1 in equation (2), we get :

**16. A homogeneous differential equation of the from dx/dt = h(x/y) can be solved by making the substitution**

(A) y = vx

(B) v = yx

(C) x = vy

(D) x = v

(A) y = vx

(B) v = yx

(C) x = vy

(D) x = v

**Solution**

Hence, the correct answer is C.

**17. Which of the following is a homogeneous differential equation?**

(A) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0

(B) (xy) dx – (x

(C) (x

(D) y

(A) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0

(B) (xy) dx – (x

^{3}+ y^{3}) dy = 0(C) (x

^{3}+ 2y^{2}) dx + 2xy dy = 0(D) y

^{2}dx + (x^{2}– xy – y^{2}) dy = 0**Solution**

^{n}F(x, y) for any non - zero constant (Î»).

Consider the equation given in alternative D :

y

^{2}dx + (x

^{2}- xy - y

^{2})dy = 0

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