# Differential Equations Reducible to Variable Separable Form

Here you will learn how to find the solution of the differential equations reducible to variable separable form with examples.

Let’s begin –

## Differential Equations Reducible to Variable Separable Form

Differential Equations of the form $$dy\over dx$$ = f(ax + by + c) can be reduce to variable separable form by the substitution ax + by + c = 0 which can be cleared by the examples given below.

Example : Solve the differential equation : $$sin^{-1}$$ $$dy\over dx$$ = x + y

Solution : We are given that,

$$sin^{-1}$$ $$dy\over dx$$ = x + y

$$\implies$$ $$dy\over dx$$ = sin(x + y)

Let x + y = v. Then

1 + $$dy\over dx$$ = $$dv\over dx$$

$$\implies$$ $$dy\over dx$$ = $$dv\over dx$$ – 1

Putting x + y = v and $$dy\over dx$$ = $$dv\over dx$$ – 1 in the given differential equation, we get

$$dv\over dx$$ – 1 = sin v

$$dv\over dx$$ = 1 + sin v

$$1\over 1 + sin v$$ dv = dx

Now, Integrating on both sides,

$$\int$$ dx = $$\int$$ $$1\over 1 + sin v$$ dv

$$\int$$ dx = $$\int$$ $$1 – sin v\over 1 – sin^2 v$$ dv

$$\int$$ dx = $$\int$$ $$1 – sin v\over cos^2 v$$ dv

$$\int$$ dx = $$\int$$ $$(sec^2 v – tan v sec v)$$ dv

x = tan v – sec v + C

$$\implies$$ x = tan (x + y) – sec (x + y) + C, which is the required solution.

Example : Solve the differential equation : $$dy\over dx$$ = cos(x + y)

Solution : We are given that,

$$dy\over dx$$ = cos(x + y)

Let x + y = v. Then

1 + $$dy\over dx$$ = $$dv\over dx$$

$$\implies$$ $$dy\over dx$$ = $$dv\over dx$$ – 1

Putting x + y = v and $$dy\over dx$$ = $$dv\over dx$$ – 1 in the given differential equation, we get

$$dv\over dx$$ – 1 = cos v

$$dv\over dx$$ = 1 + cos v

$$1\over 1 + cos v$$ dv = dx

Now, Integrating on both sides,

$$\int$$ dx = $$\int$$ $$1\over 1 + cos v$$ dv

$$\int$$ dx = $$\int$$ $$1\over 2$$ $$sec^2 v/2$$ dv

x = $$tan v/2$$ + C

$$\implies$$ x = x = $$tan (x + y)/2$$ + C, which is the required solution