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

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
(x² + xy) dy = (x² + y²) dx

Solution

The given differential equation i.e. (x² + xy) dy = (x² + y²) 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
(x² – y²) dx + 2xy dy = 0 

Solution

The given differential equation is : 
(x² – y²) 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 + v² = C/x 

⇒ x² + y² = Cx 
This is the required solution of the given differential equation.

5. Show that the given differential equation is homogeneous and solve each of them
x² (dy/dx) = x² - 2y² + 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 them
x dy -y dx = √(x² + y²)dx

Solution

x dy -y dx =  √(x² + y² )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  : 

= λ⁰. 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. 

To solve it, we make the substitution as : 

y = vx 

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 

Solution

y dx + x log(y/x) dy - 2xdy = 0 

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/v = dt/dv 
⇒ 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^x/y ) dx + e^x/y (1 - x/y)dy = 0 

Solution

Therefore, the given differential equation is a homogeneous equation. 

To solve it, we make the substitution as :  
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

Solution

(x + y)dy + (x - y)dx = 0 
⇒ (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^-11 = 2k 
⇒ log 2 + 2 × π/4 = 2k 
⇒ π/2 + log 2 = 2k 

Substituting the value of 2k in equation (2), we get : 
log(x² + y²) + 2 tan^-1(y/x) = π/2 + log 2 

This is the required solution of the given differential equation. 

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

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

Solution

 x² dy + (xy + y² )dx = 0 
⇒ x² dy = -(xy + y² )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) = C² 
⇒ C² = 1/3 
Substituting C² = 1/3 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.

13. For the differential equations find the particular solution satisfying the given condition:
[x sin² (y/x - y)] dx + xdy = 0; y = π/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 = π/4 at x = 1. 
⇒ cot (π/4) = log |C| 
⇒ 1 = log C 
⇒ C = e¹ = 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 

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 :  
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 = e¹ = 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 + y² - 2x² (dy/dx) = 0; y = 2 when x = 1 

Solution

Therefore, the given differential equation is a homogeneous equation. 

To solve it, we make the substitution as :  
y = vx 

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

Now, y = 2 at x = 1. 

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

This is the required solution of the given differential equation. 

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

Solution

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. 

17. Which of the following is a homogeneous differential equation?
(A) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0
(B) (xy) dx – (x³ + y³) dy = 0
(C) (x³ + 2y²) dx + 2xy dy = 0
(D) y² dx + (x² – xy – y²) dy = 0 

Solution

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 

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