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**CBSE Class 12**^{th}Maths## NCERT Solutions for Class 12 Maths Differential Equations

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

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.5: Ques No 1.**

(x^{2} + xy) dy = (x^{2} + y^{2}) dx

**NCERT Solutions:**

Given differential equation is

In equation (1), we see that dy/dx is in form of g(y/x), so it is a Homogeneous function of degree zero. Thus, the given differential equation is Homogeneous Differential Equation.

To solve this, we need to put y = vx …(2)

Differentiating the equation (2) with respect to x, we get

On substituting y = vx and in equation (1), we get

where C is an arbitrary constant. This is required general solution of the given differential equation.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.5: Ques No 2.**

**NCERT Solutions:**

Given differential equation is

…(1)

In equation (1), we see that dy/dx is in form of g(y/x), so it is a Homogeneous function of degree zero. Thus, the given differential equation is Homogeneous Differential Equation.

To solve this, we need to put y = vx …(2)

Differentiating the equation (2) with respect to x, we get

On substituting y = vx and in equation (1), we get

where C is an arbitrary constant. This is required general solution of the given differential equation.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.5: Ques No 3.**

(x – y) dy – (x + y) dx = 0

**NCERT Solutions:**

Given differential equation is

…(1)

In equation (1), we see that dy/dx is in form of g(y/x), so it is a Homogeneous function of degree zero. Thus, the given differential equation is Homogeneous Differential Equation.

To solve this, we need to put y = vx …(2)

Differentiating the equation (2) with respect to x, we get

On substituting y = vx and in equation (1), we get

where C is an arbitrary constant. This is required general solution of the given differential equation.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.5: Ques No 4.**

(x^{2} – y^{2}) dx + 2xy dy = 0

**NCERT Solutions:**

Given differential equation is

…(1)

To solve this, we need to put y = vx …(2)

Differentiating the equation (2) with respect to x, we get

On substituting y = vx and in equation (1), we get

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.5: Ques No 5.**

**NCERT Solutions:**

Given differential equation is

…(1)

To solve this, we need to put y = vx …(2)

Differentiating the equation (2) with respect to x, we get

On substituting y = vx and in equation (1), we get

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.5: Ques No 6.**

**NCERT Solutions:**

Given differential equation is

or

…(1)

To solve this, we need to put y = vx …(2)

Differentiating the equation (2) with respect to x, we get

On substituting y = vx and in equation (1), we get

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.5: Ques No 7.**

**NCERT Solutions:**

Given differential equation is

…(1)

To solve this, we need to put y = vx …(2)

Differentiating the equation (2) with respect to x, we get

On substituting y = vx and in equation (1), we get

Integrating, we get

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.5: Ques No 8.**

**NCERT Solutions:**

Given differential equation is

…(1)

To solve this, we need to put y = vx …(2)

Differentiating the equation (2) with respect to x, we get

On substituting y = vx and in equation (1), we get

Integrating, we get

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.5: Ques No 9.**

**NCERT Solutions:**

Given differential equation is

…(1)

To solve this, we need to put y = vx …(2)

Differentiating the equation (2) with respect to x, we get

On substituting y = vx and in equation (1), we get

Integrating, we get

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.5: Ques No 10.**

**NCERT Solutions:**

Given differential equation is

…(1)

In equation (1), we see that dx/dy is in form of g(x/y), so it is a Homogeneous function of degree zero. Thus, the given differential equation is Homogeneous Differential Equation.

To solve this, we need to put x = vy …(2)

Differentiating the equation (2) with respect to y, we get

On substituting x = vy and in equation (1), we get

Integrating, we get

For each of the differential equations in Exercises from 11 to 15, find the particular solution satisfying the given condition:

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.5: Ques No 11.**

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

**NCERT Solutions:**

Given differential equation is (x + y) dy + (x – y) dx = 0.

…(1)

To solve this, we need to put y = vx …(2)

Differentiating the equation (2) with respect to y, we get

On substituting y = vx and in equation (1), we get

Integrating, we get

…(3)

It is given that y = 1 when x = 1.

From the equation (3), we get

Substituting the value of C in equation (3), we get

This is required particular solution of the given differential equation.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.5: Ques No 12.**

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

**NCERT Solutions:**

Given differential equation is x^{2} dy + (xy + y^{2}) dx = 0.

…(1)

To solve this, we need to put y = vx …(2)

Differentiating the equation (2) with respect to y, we get

On substituting y = vx and in equation (1), we get

Integrating, we get

…(3)

where C is an arbitrary constant. This is general solution of the given differential equation.

It is given that y = 1 when x = 1.

From the equation (3), we get

Substituting the value of C in equation (3), we get

This is required particular solution of the given differential equation.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.5: Ques No 13.**

y = π/4 when x = 1

**NCERT Solutions:**

Given differential equation is

…(1)

To solve this, we need to put y = vx …(2)

Differentiating the equation (2) with respect to y, we get

On substituting y = vx and in equation (1), we get

Integrating, we get

…(3)

where C is an arbitrary constant. This is general solution of the given differential equation.

It is given that y = π/4 when x = 1.

From the equation (3), we get

Substituting the value of C in equation (3), we get

This is required particular solution of the given differential equation.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.5: Ques No 14.**

y = 0 when x = 1

**NCERT Solutions:**

Given differential equation is

…(1)

To solve this, we need to put y = vx …(2)

Differentiating the equation (2) with respect to y, we get

On substituting y = vx and in equation (1), we get

Integrating, we get

…(3)

where C is an arbitrary constant. This is general solution of the given differential equation.

It is given that y = 0 when x = 1.

From the equation (3), we get

Substituting the value of C in equation (3), we get

This is required particular solution of the given differential equation.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.5: Ques No 15.**

y = 2 when x = 1

**NCERT Solutions:**

Given differential equation is

…(1)

To solve this, we need to put y = vx …(2)

Differentiating the equation (2) with respect to y, we get

On substituting y = vx and in equation (1), we get

Integrating, we get

…(3)

where C is an arbitrary constant. This is general solution of the given differential equation.

It is given that y = 2 when x = 1.

From the equation (3), we get

Substituting the value of C in equation (3), we get

This is required particular solution of the given differential equation.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.5: Ques No 16.**

A homogeneous differential equation of the from can be solved by making the substitution.

(A) y = vx

(B) v = yx

(C) x = vy

(D) x = v

**NCERT Solutions:**

To solve the homogeneous differential equation of the from , we need to put x = vy.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.5: Ques No 17.**

Which of the following is a homogeneous differential equation?

(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

**NCERT Solutions:**