Question

Form the differential equation having $y={\left({\sin }^{-1}x\right)}^2+A{\cos }^{-1}x+B$, where A and B are arbitrary constants, as its general solution.

Answer 1

We have two constnts in the solution, so we will differentiate both sides twice and eliminate the constants A and B.

$$\begin{gathered} y={\left({\sin }^{-1}x\right)}^2+A{\cos }^{-1}x+B \\ \Rightarrow \frac{dy}{dx}=\frac{2{\sin }^{-1}x}{\sqrt{1-x^2}}-\frac{A}{\sqrt{1-x^2}} \\ \Rightarrow \left(\sqrt{1-x^2}\right)\frac{dy}{dx}=2{\sin }^{-1}x-A \end{gathered}$$
$$\begin{gathered} \Rightarrow \left(\sqrt{1-x^2}\right)\frac{d^{2}y}{d{x}^2}+\frac{\left(-2x\right)}{2\sqrt{1-x^2}}\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}} \\ \Rightarrow \left(\sqrt{1-x^2}\right)\frac{d^{2}y}{d{x}^2}-\frac{x}{\sqrt{1-x^2}}\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}} \\ \Rightarrow \left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=2 \\ \Rightarrow \left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-2=0 \end{gathered}$$