Question

\[\sqrt{1 - x^4} dy = x\ dx\]

Answer 1

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