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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 constant, as its general solution.

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

Answers (1)

Given:

y=(\arcsin x)^{2}+A \arccos x+B

To find: Solution of the differential equation

Differentiating on both the sides,

\frac{dy}{dx}=2 \arcsin x\frac{1}{\sqrt{1-x^{2}}}-A\frac{1}{\sqrt{1-x^{2}}}\\ \sqrt{1-x^{2}}\frac{dy}{dx}=2 \arcsin x-A\\ Formula: \frac{d}{dx}\left ( \arcsin x \right )=\frac{1}{\sqrt{1-x^{2}}}\\\frac{d}{dx}\left ( \arccos x \right )=-\frac{1}{\sqrt{1-x^{2}}}

Differntiate again on both the sides

\sqrt{1-x^{2}}\frac{d^{2}y}{dx^{2}}-\frac{dy}{dx}\frac{x}{\sqrt{1-x^{2}}}=2\frac{1}{\sqrt{1-x^{2}}}\\ \left ( 1-x^{2} \right )\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}=2\\ Formula: \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\\

Hence the solution is
\left ( 1-x^{2} \right )\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}-2=0\\

Posted by

infoexpert24

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