The solution of the differential equation $x\frac{dy}{dx}+y=x^3{y}^6$ , is 

The given differential equation can be written as

$\frac{1}{y^6}\frac{dy}{dx}+\frac{1}{x{y}^5}=x^2$

Let $y^{-5}=0$ , Then 

$-5y^{-6}\frac{dy}{dx}=\frac{dv}{dx}\Rightarrow y^{-6}\frac{dy}{dx}=-\frac{1}{5}\frac{dv}{dx}$

Substituting these values in the given differential equation, we get

$-\frac{1}{5}\frac{dv}{dx}+\frac{1}{x}v=x^2$

$\Rightarrow \frac{dv}{dx}-\frac{5}{x}v=-5x^2$          …(i)

This is the standard form of the linear differential equation having integrating factor

$\text{ I. F. }=e^{\int \frac{5}{x}dx}=e^{-5\log x}=\frac{1}{x^5}$

Multiplying both sides of (i) by I.F. and integrating w.r.t. x, we get 

$\begin{array}{l}v\frac{1}{x^5}=\int -5x^2\cdot \frac{1}{x^5}dx\\ \Rightarrow \frac{v}{x^5}=\frac{5}{2}{x}^{-2}+C\end{array}$

$\Rightarrow y^{-5}{x}^5=\frac{5}{2}{x}^{-2}+C$  , which is the required solution.