**Chapter 6 Differential Equations Ex 6.3**

## Chapter 6 Differential Equations Ex 6.3

**Question 1.In each of the following examples verify that the given expression is a solution of the corresponding differential equation.**

Solution:

Solution:

xy = log y + c

Differentiating w.r.t. x, we get

Hence, xy = log y + c is a solution of the D.E.

Differentiating w.r.t. x, we get

Differentiating again w.r.t. x, we get

Differentiating w.r.t. x, we get

Differentiating again w.r.t. x, we get

Solution:

This is the general solution.

Multiplying throughout by 4, this becomes

Solution:

Solution:

Solution:

This is the general solution.

**Question 3.For each of the following differential equations, find the particular solution satisfying the given condition:**

When x = 2, y = 0, we have

(1 + 4)(1 – 0) = c

∴ c = 5

**Question 4.Reduce each of the following differential equations to the variable separable form and hence solve:**

**Solution:**

Solution:

Solution:

Integrating both sides, we get

Integrating both sides, we get

∫dx = ∫seu du

∴ x = tan u + c

∴ x = tan(x – 2y) + c

This is the general solution.

**(v) (2x – 2y + 3) dx – (x – y + 1) dy = 0, when x = 0, y = 1Solution:**

(2x – 2y + 3) dx – (x – y + 1) dy = 0

∴ (x – y + 1) dy = (2x – 2y + 3) dx

∴ u – log|u + 2| = -x + c

∴ x – y – log|x – y + 2| = -x + c

∴ (2x – y) – log|x – y + 2| = c

This is the general solution.

Now, y = 1, when x = 0.

∴ (0 – 1) – log|0 – 1 + 2| = c

∴ -1 – o = c

∴ c = -1

∴ the particular solution is

(2x – y) – log|x – y + 2| = -1

∴ (2x – y) – log|x – y + 2| + 1 = 0