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# 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.

Toolbox:
• A differential equation is a linear differential equation if it is expressible in the form : $P_0 \Large\frac{d^ny}{dx^n}$$+P_1 \Large\frac{d^{n-1}y}{dx^{n-1}}$$+P_2\Large \frac{d^{n-2}y}{dx^{n-2}}$$+...P_n y=0 Where P_0,P_1.... are constants or functions of the independent of variable x • \large\frac{d}{dx}$$(\sin ^{-1}x)=\large\frac{1}{\sqrt {1-x^2}}$
• $\large\frac{d}{dx}$$(uv)=\large\frac{du}{dx}.$$v+\large\frac{dv}{dx}.$$u Given y= (\sin ^{-1}x)^2+A\cos ^{-1}x+B Let us first differentiate with respect to x on both sides \large\frac{dy}{dx}=\frac{2\sin ^{-1}x}{\sqrt {1-x^2}}$$+A\bigg(\large\frac{-1}{\sqrt {1-x^2}}\bigg)+0$
$\large\frac{dy}{dx}=\frac{2\sin ^{-1}x}{\sqrt {1-x^2}}-\frac{A}{\sqrt {1-x^2}}$
This can be written as
$\sqrt {1-x^2}.\large\frac{dy}{dx}$$=2 \sin ^{-1}x-A Now again differentiating on both sides w.r.t x Since \sqrt {1-x^2}.\large\frac{dy}{dx} are two functions in the product form, we can apply the product rule \large\frac{d}{dx}$$(uv)=u.\large\frac{d}{dx}$$(v)+v.\large\frac{d}{dx}$$(u)$
Here let $u=\sqrt {1-x^2},$ hence $\large\frac{d}{dx}$$(v)=\large\frac{1}{2 \sqrt {1-x^2}}$$(-2x)$
$v=\large\frac{dy}{dx}\qquad \frac{d}{dx}$$(v)=\large\frac{d^2y}{dx^2} \large\frac{d}{dx}$$(\sin ^{-1}x)=\large\frac{1}{\sqrt {1-x^2}}$
$\sqrt {1-x^2}.\large \frac{d^2y}{dx^2}+\frac{dy}{dx}.\frac{-2x}{2\sqrt {1-x^2}}$$=2 \large\frac{1}{\sqrt {1-x^2}}-0 \sqrt {1-x^2}.\large \frac{d^2y}{dx^2}-\frac{x}{\sqrt {1-x^2}}.\frac{dy}{dx}$$= \large\frac{2}{\sqrt {1-x^2}}$
$\Large\frac{(1-x^2)\bigg(\frac{d^2y}{dx^2}\bigg)-x\bigg(\frac{dy}{dx}\bigg)}{\sqrt {1-x^2}}=\frac{2}{\sqrt {1-x^2}}$
=>$(1-x^2)\large\frac{d^2y}{dx^2}$$-x \large\frac{dy}{dx}$$-2=0$