# Differential Equations Reducible to Variable Separable Method

## Differential Equations Reducible to Variable Separable Method

Differential equation of the first order cannot be solved directly by variable separable method. But by some substitution, we can reduce it to a differential equation with separable variable. Let the differential equation is of the form

$$\frac{dy}{dx}=f\left( ax+by+c \right)$$,

Can be reduced to variable separable form by the substitution ax + by + c = z

∴ $$a+b\frac{dy}{dx}=\frac{dz}{dx}$$,

∴ $$\left( \frac{dz}{dx}-a \right)\frac{1}{b}=f(z)$$,

$$\frac{dz}{dx}=a+bf(z)$$,

Now, apply variable separable method.

Example: Find the solution of the differential equation $$\frac{dy}{dx}=\sin \left( x+y \right)+\cos \left( x+y \right)$$.

Solution: We have, $$\frac{dy}{dx}=\sin \left( x+y \right)+\cos \left( x+y \right)$$ … 1

Let z = x + y → $$\frac{dz}{dx}=1+\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{dz}{dx}-1$$,

From eq. (i),

$$\frac{dz}{dx}-1$$ = sinz + cosz

$$\Rightarrow \frac{dz}{\sin z+\cos z+1}=dx$$ (variables are separated)

On integrating both sides, we get

$$\int{\frac{dx}{\sin z+\cos z+1}}=\int{1.dx+C}$$,

$$\Rightarrow \int{\frac{dz}{\frac{2\tan \frac{z}{2}}{1+{{\tan }^{2}}\frac{z}{2}}+\frac{1-{{\tan }^{2}}\frac{z}{2}}{1+{{\tan }^{2}}\frac{z}{2}}+1}=x+C}$$,

$$\Rightarrow \int{\frac{{{\sec }^{2}}\frac{z}{2}dz}{2\tan \frac{z}{2}+2}}=x+C$$,

$$\Rightarrow \int{\frac{dt}{t+1}=x+C}$$, Where t = $$\tan \frac{z}{2}$$,

log |t + 1| = x + C

$$\Rightarrow \log \left| \tan \frac{x+y}{2}+1 \right|=x+C$$.

This is the required general solution.