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Q.
If the curve $y=f(x)$ passing through the point $(1,2)$ and satisfies the differential equation $x d y+\left(y+x^{3} y^{2}\right)$ $d x=0$, then
Differential Equations
Solution:
$x d y+\left(y+x^{3} y^{2}\right) d x=0$
$ \Rightarrow x d y+y d x=-x^{3} y^{2} d x$
$\Rightarrow \frac{x d y+y d x}{x^{2} y^{2}}=-x d x $
$\Rightarrow \frac{d(x y)}{(x y)^{2}}=-x d x$
Integrating, we get
$-\frac{1}{x y}=-\frac{x^{2}}{2}+c ...$(1)
Using $(1,2)$ in $(1)$, we get
$-\frac{1}{2}=-\frac{1}{2}+c \Rightarrow c=0$
$\therefore -\frac{1}{x y}=-\frac{x^{2}}{2} $
$\Rightarrow y=\frac{2}{x^{3}}$ or $x^{3} y=2$
is the required curve.