Question

$\sqrt{1-x^4}dy=xdx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \sqrt{1-x^4}dy=xdx \\ \Rightarrow dy=\frac{x}{\sqrt{1-x^4}}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int dy=\int \frac{x}{\sqrt{1-x^4}}dx \\ \Rightarrow y=\int \frac{x}{\sqrt{1-x^4}}dx \\ \mathrm{Putting}{x}^2=t \\ \Rightarrow 2xdx=dt \\ \therefore y=\frac{1}{2}\int \frac{dt}{\sqrt{1-t^2}} \\ =\frac{{\sin }^{-1}t}{2}+C \\ =\frac{1}{2}{\sin }^{-1}\left(x^2\right)+C \\ \mathrm{Hence},y=\frac{1}{2}{\sin }^{-1}\left(x^2\right)+C\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}. \end{gathered}$$