Question:

For the differential equations , find the general solution:\(\frac{dy}{dx}=sin^{-1}x\)

Updated On: Oct 3, 2023
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Solution and Explanation

The given differential equation is:
\(\frac{dy}{dx}=sin^{-1}x\)
\(⇒dy=sin^{-1}xdx\)
Integrating both sides,we get:
\(∫dy=∫sin^{-1}xdx\)
\(⇒y=∫(sin^{-1}x.1)dx\)
\(⇒y=sin^{-1}x.∫(1)dx-∫[(\frac{d}{dx}(sin^{-1}x).∫(1)dx)]dx\)
\(⇒y=sin^{-1}x.x-∫(\frac{1}{\sqrt{1-x^2}}.x)dx\)
\(⇒y=xsin^{-1}x+∫\frac{-x}{\sqrt{1-x^2}}dx...(1)\)
Let \(1-x^2=t.\)
\(⇒\frac{d}{dx}(1-x^2)=\frac{dt}{dx}\)
\(⇒-2x=\frac{dt}{dx}\)
\(⇒xdx=\frac{-1}{2}dt\)
Substituting this value in equation(1),we get:
\(y=xsin^{-1}x+∫\frac{1}{2}\sqrt{t}dt\)
\(⇒y=xsin^{-1}x+\frac{1}{2}.∫(t)-\frac{1}{2}dt\)
\(⇒y=xsin^{-1}x+\sqrt{t}+C\)
\(⇒y=xsin^{-1}x+\sqrt{1-x^2}+C\)
This is the required general solution of the given differential equation.
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