# Differential Equations in Variable Separable Form

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

Let’s begin –

## Differential Equations in Variable Separable Form

If the differential equation can be put in the form f(x) dx = g(y) dy, we say that the variables are seperable and such equations can be solved by integrating on both sides. The solution is given by

$$\int$$ f(x) dx = $$\int$$ g(y) dy + C, where C is an arbitrary constant

Note : There is no need of introducing arbitrary constants of integration on both sides as they can be combined together to give just one arbitrary constant.

Example : Solve the differential equation : (x + 1)$$dy\over dx$$ = 2xy

Solution : We have,

(x + 1)$$dy\over dx$$ = 2xy

$$\implies$$ (x + 1)dy = 2xy dx

$$\implies$$ $$dy\over y$$ = $$2x\over x + 1$$ dx

Now, integrating on both sides,

$$\int$$ $$1\over y$$ dy = 2 $$\int$$ $$x\over x + 1$$ dx

$$\implies$$ $$1\over y$$ dy = 2 $$\int$$ $$x + 1 – 1\over x + 1$$ dx

$$\implies$$ $$1\over y$$ dy = 2 $$\int$$ $$1 – {1\over x + 1}$$ dx

log y = 2{x – log| x + 1 |} + C, which is the solution of the given differential equation.

Example : Solve the differential equation : cos x(1 + cos y) dx  – sin y(1 + sin x)dy = 0

Solution : We have,

cos x(1 + cos y) dx  – sin y(1 + sin x)dy = 0

$$\implies$$ $$cos x\over 1 + sin x$$ dx – $$sin y\over 1 + cos y$$ dy = 0

$$\implies$$ $$cos x\over 1 + sin x$$ dx + $$-sin y\over 1 + cos y$$ dy = 0

Now, integrating on both sides,

$$\int$$ $$cos x\over 1 + sin x$$ dx + $$\int$$ $$-sin y\over 1 + cos y$$ dy = 0

$$\implies$$ log|1 + sin x| + log|1 + cos y| = log C

$$\implies$$ log{|1 + sin x|.|1 + cos y|} = log C

|1 + sin x|.|1 + cos y| = C

Hence, (1 + sin x)(1 + cos y) = C is the solution of the given differential equation.