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Find the general solution of the differential equation

Question:

Find the general solution of the differential equation $\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0$

Solution:

$\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0$

$\Rightarrow \frac{d y}{d x}=-\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}$

$\Rightarrow \frac{d y}{\sqrt{1-y^{2}}}=\frac{-d x}{\sqrt{1-x^{2}}}$

Integrating both sides, we get:

$\sin ^{-1} y=-\sin ^{-1} x+\mathrm{C}$

$\Rightarrow \sin ^{-1} x+\sin ^{-1} y=\mathrm{C}$

 

 

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