The solution of \(\frac{\mathrm{d} x}{\mathrm{~d} y}+\frac{x}{y}=x^2\) is

\(\frac{\mathrm{d} x}{\mathrm{~d} y}+\frac{x}{y}=x^2\)
\(\therefore \quad \frac{1}{x^2} \frac{\mathrm{d} x}{\mathrm{~d} y}+\frac{1}{x y}=1.. (i)\)
Let \(\frac{1}{x}=\mathrm{t}\)
Differentiating w.r.t. \(y\), we get
\(\begin{array}{ll}
& \frac{-1}{x^2} \frac{\mathrm{d} x}{\mathrm{~d} y}=\frac{\mathrm{dt}}{\mathrm{d} y} \Rightarrow \frac{1}{x^2} \frac{\mathrm{d} x}{\mathrm{~d} y}=\frac{-\mathrm{dt}}{\mathrm{d} y} \\
\therefore \quad & \text { (i) } \Rightarrow \frac{-\mathrm{dt}}{\mathrm{d} y}+\frac{\mathrm{t}}{y}=1 \\
\therefore \quad & \frac{\mathrm{dt}}{\mathrm{d} y}-\frac{\mathrm{t}}{y}=-1 \\
\therefore & \text { I.F. }=\mathrm{e}^{\int \frac{-1}{y} \mathrm{~d}}=\mathrm{e}^{\log \left(\frac{1}{y}\right)}=\frac{1}{y}
\end{array}\)
\(\therefore \quad\) The solution of the given equation is
\(\begin{aligned}
\mathrm{t}(\mathrm{I} . \mathrm{F}) & =\int(-1)(\text { I.F. }) \mathrm{d} y+\mathrm{c} \\
\mathrm{t}\left(\frac{1}{y}\right) & =\int \frac{-1}{y} \mathrm{~d} y+\mathrm{c} \\
\therefore \quad \frac{\mathrm{t}}{y} & =-\log y+\mathrm{c} \\
\therefore \quad \frac{1}{x y} & =-\log y+\mathrm{c} \\
\therefore \quad \frac{1}{x} & =\mathrm{c} y-y \log y
\end{aligned}\)