The solution of \(r d x+\left(x-r^{2}\right) d r=0\) is

Given \(\mathrm{rdx}+\left(\mathrm{x}-\mathrm{r}^{2}\right) \mathrm{dr}=0 \Rightarrow \mathrm{rdx}=-\left(\mathrm{x}-\mathrm{r}^{2}\right) \mathrm{dr}\)
\(\begin{array}{l}
\therefore \mathrm{r} \frac{\mathrm{dx}}{\mathrm{dr}}=\mathrm{r}^{2}-\mathrm{x} \Rightarrow \mathrm{r} \frac{\mathrm{d} \mathrm{x}}{\mathrm{dr}}+\mathrm{x}=\mathrm{r}^{2} \\
\therefore \frac{\mathrm{d} \mathrm{x}}{\mathrm{dr}}+\frac{\mathrm{x}}{\mathrm{r}}=\mathrm{r} \\
\quad \text { I.F. }=\mathrm{e}^{\int \frac{1}{\mathrm{r}} \mathrm{dr}}=\mathrm{e}^{\log \mathrm{r}}=\mathrm{r} \\
\text { Solution is } \mathrm{x} \cdot \mathrm{r}=\int \mathrm{r} \cdot \mathrm{r} \mathrm{dr}+\mathrm{c} \\
\quad \mathrm{xr}=\frac{\mathrm{r}^{3}}{3}+\mathrm{c}
\end{array}\)