Class 12 NCERT Solutions- Mathematics Part ii – Chapter 9– Differential Equations Exercise 9.4

NCERT solutions Class 12 Chapter 9 Exercise 9.4 consists of 16 questions that impart a clear understanding of Differential Equations. Learn about the concept used and the solution to Chapter 9– Integrals Exercise 9.4 in this article.

Question 1. dy/dx = (x² + y²) / (x² + xy)

Solution:

Given equation:

Thus, the given equation is homogenous.

Let y = vx

(x - y)² = Cxe^y/x

Question 2.

Solution:

Given equation can be rearranged as:

Thus, the given equation is homogenous.

Let y = vx.

y = xlog|x| + Cx

Question 3. (x-y)dy - (x+y)dx = 0

Solution:

Given equation can be rearranged as:

Thus, the given equation is homogenous.

Let y = vx.

tan^-1(y/x) = 1/2[log(x² + y²)] + C

Question 4. (x²-y²)dx + 2xy dy = 0

Solution:

Given equation can be rearranged as:

Thus, the given equation is homogenous.

Let y = vx.

⇒ x² + y² = Cx

Question 5.

Solution:

Given equation can be rearranged as:

Thus, the given equation is homogenous.

Let y = vx.

Question 6.

Solution:

Given equation can be rearranged as:

Thus, the given equation is homogenous.

Let y = vx.

Question 7.

Solution:

Given equation can be rearranged as:

Thus, the given equation is homogenous.

Let y = vx.

Question 8.

Solution:

Given equation can be rearranged as:

Thus, the given equation is homogenous.

Let y = vx.

Question 9.

Solution:

Given equation can be rearranged as:

Thus, the given equation is homogenous.

Let y = vx.

Question 10.

Solution:

Given equation can be rearranged as:

Thus, the given equation is homogenous.

Let y = vx.

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

Question 11. (x + y)dy + (x – y)dx = 0; y = 1 when x = 1

Solution:

Given equation can be rearranged as:

Thus, the given equation is homogenous.

Let y = vx.

Since y = 1 when x = 1, we have:

⇒ log 2 + 2 tan^-11 = 2k

⇒ π/2 + log 2 = 2k

Thus, is the required equation.

Question 12. x² dy + (xy + y²) dx = 0; y = 1 when x = 1

Solution:

Given equation can be rearranged as:

Thus, the given equation is homogenous.

Let y = vx.

Since y = 1 when x = 1, we have:

C² = 1/3

Thus, y + 2x = 3x²y is the required equation.

Question 13.

Solution:

Given equation can be rearranged as:

Thus, the given equation is homogenous.

Let y = vx.

Now, y = π/4 when x = 1, we have:

1 = log C or C = e.

Thus, is the required equation.

Question 14. ; y = 0 when x = 1.

Solution:

Given equation can be rearranged as:

Thus, the given equation is homogenous.

Let y = vx.

Now, y = 0 when x = 1, we have:

cos(0) = log C

⇒ C = e

Thus, is the required solution.

Question 15. ; y = 2 when x = 1

Solution:

Given equation can be rearranged as:

Thus, the given equation is homogenous.

Let y = vx.

Now, y = 2 when x = 1, we have:

-1 = log(1) + C

C = -1

Thus,is the required equation.

Question 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

Solution:

Option (C) is Correct

Question 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:

Option (D) is Correct