Question

$\frac{dy}{dx}=\frac{1+y^2}{y^3}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}=\frac{1+y^2}{y^3} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{dx}{dy}=\frac{y^3}{1+y^2} \\ \Rightarrow dx=\frac{y^3}{1+y^2}dy \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int dx=\int \frac{y^3}{1+y^2}dy \\ \Rightarrow x=\int \frac{y+y^3-y}{1+y^2}dy \\ \Rightarrow x=\int \frac{\left(1+y^2\right)y-y}{1+y^2}dy \\ \Rightarrow x=\int ydy-\int \frac{y}{1+y^2}dy \\ \Rightarrow x=\frac{y^2}{2}-\int \frac{y}{1+y^2}dy \\ \mathrm{Putting}1+y^2=t\mathrm{we}\mathrm{get} \\ 2ydy=dt \\ \therefore x=\frac{y^2}{2}-\frac{1}{2}\int \frac{1}{t}dt \\ \Rightarrow x=\frac{y^2}{2}-\frac{1}{2}\log \left|t\right|+C \\ \Rightarrow x=\frac{y^2}{2}-\frac{1}{2}\log \left|1+y^2\right|+C\left(\because t=1+y^2\right) \\ \mathrm{Hence},x=\frac{y^2}{2}-\frac{1}{2}\log \left|1+y^2\right|+C\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$