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Mathematics Question on Differential equations

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

Answer

The given differential equation is:
dydx=sin−1x\frac{dy}{dx}=sin^{-1}x
⇒dy=sin−1xdx⇒dy=sin^{-1}xdx
Integrating both sides,we get:
∫dy=∫sin−1xdx∫dy=∫sin^{-1}xdx
⇒y=∫(sin−1x.1)dx⇒y=∫(sin^{-1}x.1)dx
⇒y=sin−1x.∫(1)dx−∫[(ddx(sin−1x).∫(1)dx)]dx⇒y=sin^{-1}x.∫(1)dx-∫[(\frac{d}{dx}(sin^{-1}x).∫(1)dx)]dx
⇒y=sin−1x.x−∫(11−x2.x)dx⇒y=sin^{-1}x.x-∫(\frac{1}{\sqrt{1-x^2}}.x)dx
⇒y=xsin−1x+∫−x1−x2dx...(1)⇒y=xsin^{-1}x+∫\frac{-x}{\sqrt{1-x^2}}dx...(1)
Let 1−x2=t.1-x^2=t.
⇒ddx(1−x2)=dtdx⇒\frac{d}{dx}(1-x^2)=\frac{dt}{dx}
⇒−2x=dtdx⇒-2x=\frac{dt}{dx}
⇒xdx=−12dt⇒xdx=\frac{-1}{2}dt
Substituting this value in equation(1),we get:
y=xsin−1x+∫12tdty=xsin^{-1}x+∫\frac{1}{2}\sqrt{t}dt
⇒y=xsin−1x+12.∫(t)−12dt⇒y=xsin^{-1}x+\frac{1}{2}.∫(t)-\frac{1}{2}dt
⇒y=xsin−1x+t+C⇒y=xsin^{-1}x+\sqrt{t}+C
⇒y=xsin−1x+1−x2+C⇒y=xsin^{-1}x+\sqrt{1-x^2}+C
This is the required general solution of the given differential equation.