Question

Form the differential equation having $$y={({\sin }^{-1}x)}^2+A{\cos }^{-1}x+B$$, 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.
$$y={({\sin }^{-1}x)}^2+A{\cos }^{-1}x+B$$
$$\Rightarrow \frac{dy}{dx}=\frac{2{\sin }^{-1}x}{\sqrt{1-x^2}}-\frac{A}{\sqrt{1-x^2}}$$
$$\Rightarrow \left(\sqrt{1-x^2}\right)\frac{dy}{dx}=2{\sin }^{-1}x-A$$
$$\Rightarrow \left(\sqrt{1-x^2}\right)\frac{d^{2}y}{d{x}^2}+\frac{(-2x)}{2\sqrt{1-x^2}}\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}$$
$$\Rightarrow \left(\sqrt{1-x^2}\right)\frac{d^{2}y}{d{x}^2}-\frac{x}{\sqrt{1-x^2}}\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}$$
$$\Rightarrow (1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=2$$
$$\Rightarrow (1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-2=0$$