Solve the following differential equation 1-x^4 dy=x dx - Vidyadip

Question

Solve the following differential equation
$\sqrt{1-\text{x}^4}\text{dy}=\text{x dx}$

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Answer

We have,
$\sqrt{1-\text{x}^4}\text{dy}=\text{x dx}$
$\Rightarrow\text{dy}=\frac{\text{x}}{\sqrt{1-\text{x}^4}}\ \text{dx}$
Integrating both sides, we get
$\int\text{dy}=\int\frac{\text{x}}{\sqrt{1-\text{x}^4}}\text{ dx}$
$\Rightarrow\text{y}=\int\frac{\text{x}}{\sqrt{1-\text{x}^4}}\ \text{dx}$
Putting $x^2= t$
$\Rightarrow2\text{x dx}=\text{dt}$
$\therefore\text{y}=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{1-\text{t}^2}}$
$=\frac{\sin^{-1}}{2}+\text{C}$
$=\frac{1}{2}\sin^{-1}(\text{x}^2)+\text{C}$
hence, $\text{y}=\frac{1}{2}\sin^{-1}(\text{x}^2)+\text{C}$ is the solution to the given differential equation.

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