# Differential Equation Reducible to Linear Form

## Differential Equation Reducible to Linear Form

An equation of the form $$f’\left( y \right)\frac{dy}{dx}+Pf\left( y \right)=Q$$… (i)

Where P and Q are constants or functions of x alone, can be reduced to linear form by putting f (y) = v.

$$f’\left( y \right)\frac{dy}{dx}=\frac{dv}{dx}$$,

Then, Equation (i) becomes $$\frac{dv}{dx}+Pv=Q$$. Which is linear in v and x.

Example: Find the solution of {xy³ (1 + cosx) – y} dx + x dy = 0.

Solution: Given that {xy³ (1 + cosx) – y} dx + x dy = 0.

The given equation can be written as $$\frac{dy}{dx}+{{y}^{3}}\left( 1+\cos x \right)-\frac{y}{x}=0$$.

i.e, $$\frac{1}{{{y}^{3}}}\frac{dy}{dx}-\frac{1}{{{y}^{2}}x}=-\left( 1+\cos x \right)$$.

Using the transformation $$\frac{-1}{{{y}^{2}}}=u$$,

Differentiation with respect ‘u’ We get

$$\frac{2}{{{y}^{3}}}dy=du$$.

The above equation reduces to $$\frac{1}{2}\frac{du}{dx}+\frac{u}{x}=-\left( 1+\cos x \right)$$.

Whose $$IF={{e}^{\int{\frac{2}{x}}dx}}={{e}^{2\ln x}}={{x}^{2}}$$.

Hence, the solution of the given differential equation is given by ux² = -2 ∫x² (1 + cosx) dx,

i.e, $$\frac{{{x}^{2}}}{2{{y}^{2}}}=\int{{{x}^{2}}}dx+\int{{{x}^{2}}}\cos x\,dx=\frac{{{x}^{3}}}{3}+{{x}^{2}}\sin x-\int{2x}\sin x\,dx$$.

$$=\frac{{{x}^{3}}}{3}+{{x}^{2}}\sin x+2x\,cosx-2\int{\cos x}+C$$.

$$=\frac{{{x}^{3}}}{3}+{{x}^{2}}\sin x+2x\,cosx-2\sin x+C$$.