Question

Solve: $y(y\log x-1)dx=xdy$

Answer 1

$y(y\log x-1)dx=xdy$

$y^2\log x-y=x\frac{dy}{dx}$

Dividing by $x{y}^2$, we get

$\frac{1}{y^2}\frac{dy}{dx}=\frac{\log x}{x}-\frac{1}{xy}$

$\frac{1}{y^2}\frac{dy}{dx}+\frac{1}{xy}=\frac{\log x}{x}$

Now., let $\frac{-1}{y}=z$

$+\frac{1}{y^2}dy=dz$

$\frac{dz}{dx}-\frac{z}{x}=\frac{\log x}{x}$

Now,$IF=e^{\int \frac{-1}{x}dx}=e^{-\log x}=\frac{1}{x}$

Now, multipying with IF we get,

$\frac{d}{dx}(\frac{1}{x}\cdot z)=\frac{\log x}{x^2}$

On RHS, let $\log x=t$

$\frac{1}{x}dx=dt$

$x=e^t$

$d\left(\frac{1}{x}z\right)=e^{-t}tdt$

Now by integrating,we get

$\frac{1}{x}z=\int e^{-t}tdt$

Using Product formula on RHS, we get

$\frac{z}{x}=t\int e^{-t}-\int \left(\right(\int e^{-t}\left)\frac{d}{dt}\right(t\left)\right)dt$

$\frac{z}{x}=-t{e}^{-t}+\int e^{-t}dt$

$\frac{z}{x}=e^{-t}(-t-1)+C$

$\frac{z}{x}=-e^{-t}(t+1)+C$

As, $z=\frac{-1}{y},t=\log x$

$\frac{1}{xy}=e^{-\log x}(\log x+1)+C$

$\frac{1}{xy}=\frac{1}{x}(\log x+1)+C$