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.