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Question

Mathematics Question on Differential equations

Find the general solution of the differential equation:
dydx+1−y21−x2=0\frac {dy}{dx}+\sqrt {\frac {1-y^2}{1-x^2}}=0

Answer

dydx+1−y21−x2=0\frac {dy}{dx}+\sqrt {\frac {1-y^2}{1-x^2}}=0

⇒dydx=−1−y21−x2\frac {dy}{dx}=-\sqrt {\frac {1-y^2}{1-x^2}}

⇒dy1−y2=−dx1−x2\frac {dy}{\sqrt {1-y^2}}=-\frac {dx}{\sqrt {1-x^2}}

Integrating both sides, we get:

sin−1y=−sin−1x+Csin^{-1}y=-sin^{-1}x+C

⇒sin−1x+sin−1y=Csin^{-1}x+sin^{-1}y=C