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Q.
The curve satisfying the equation $\frac{d y}{d x}=\frac{y\left(x+y^{3}\right)}{x\left(y^{3}-x\right)}$ and passing through the point $(4,-2)$ is
Differential Equations
Solution:
$\left(x y^{3}-x^{2}\right) d y-\left(x y+y^{4}\right) d x=0$
$\Rightarrow y^{3}(x d y-y d x)-x(x d y+y d x)=0$
$\Rightarrow x^{2} y^{3} \frac{(x d y-y d x)}{x^{2}}-x(x d y+y d x)=0$
$\Rightarrow x^{2} y^{3} d\left(\frac{y}{x}\right)-x d(x y)=0$
$\Rightarrow $ Dividing by $x^{3} y^{2},$ we get
$\frac{y}{x} d\left(\frac{y}{x}\right)-\frac{d(x y)}{x^{2} y^{2}}=0$
Now, integrating $\frac{1}{2}\left(\frac{y}{x}\right)^{2}+\frac{1}{x y}=c$
It passes through the point $(4,-2) $ .
$\Rightarrow \frac{1}{8}-\frac{1}{8}=c $
$ \Rightarrow c=0$
$\therefore y^{3}=-2 x$