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.