[linear differential equation in x]
$\therefore \quad {\rm{IF}} = {e^{\int {\frac{1}{{{y^2}}}dy} }} = {e^{ - \frac{1}{y}}}$
Complete solution is
$x \cdot e^{-\frac{1}{y}}=\int \frac{1}{y^{3}} \cdot e^{-\frac{1}{y}} d y$
$\Rightarrow \quad x \cdot e^{-\frac{1}{y}}=\int \frac{1}{y} \cdot \frac{1}{y^{2}} \cdot e^{-\frac{1}{y}} d y$
Put $\quad-\frac{1}{y}=t \Rightarrow \frac{1}{y^{2}} d y=d t$
$\Rightarrow \quad x e^{-\frac{1}{y}}=\int-t \cdot e^{t} d t$
$\Rightarrow \quad x e^{-\frac{1}{y}}=-\left\{t \cdot e^{t}-\int 1 \cdot e^{t} d t\right\}+C$
$\Rightarrow \quad x e^{-\frac{1}{y}}=-t e^{t}+e^{t}+C$
$ \Rightarrow \quad x{e^{ - \frac{1}{y}}} = \frac{1}{y} \cdot {e^{ - \frac{1}{y}}} + {e^{ - \frac{1}{y}}} + C$
$\therefore \quad y(1)=1$
$\therefore \quad e^{-1}=e^{-1}+e^{-1}+C$
$\Rightarrow \quad C=-\frac{1}{e}$
$\therefore \quad x e^{-\frac{1}{y}}=\frac{1}{y} \cdot e^{-\frac{1}{y}}+e^{-\frac{1}{y}}-\frac{1}{e}$
$\Rightarrow \quad x=\frac{1}{y}+1-\frac{1}{e} \cdot e^{\frac{1}{y}}$
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