The solution of the differential equation $x \frac{d y}{d x}+y=x^3 y^6$, is
Answer & explanation
Correct answer: option 2
The given differential equation can be written as
$\frac{1}{y^6} \frac{d y}{d x}+\frac{1}{x y^5}=x^2$
Let $y^{-5}=v$. Then,
$-5 y^{-6} \frac{d y}{d x}=\frac{d v}{d x} \Rightarrow y^{-6} \frac{d y}{d x}=-\frac{1}{5} \frac{d v}{d x}$
Substituting these values in the given differential equation, we get
$-\frac{1}{5} \frac{d v}{d x}+\frac{1}{x} v=x^2$
$\Rightarrow \frac{d v}{d x}-\frac{5}{x} v=-5 x^2$ .......(i)
This is the standard form of the linear differential equation having integrating factor
Integrating factor = $e^{\int-\frac{5}{x} d x}=e^{-5 \log x}=\frac{1}{x^5}$
Multiplying both sides of (i) by integrating factor and integrating w.r.t. $x$, we get
$v \frac{1}{x^5}=\int-5 x^2 . \frac{1}{x^5} d x$
$\Rightarrow \frac{v}{x^5}=\frac{5}{2} x^{-2}+C$
$\Rightarrow y^{-5} x^5=\frac{5}{2} x^{-2}+C$, which is the required solution.