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  • Find the general solution of the differential equation Differential Equations Miscellaneous Exercise 6.

Q6.    Find the general solution of the differential equation \frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1-x^2}} = 0

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Given equation is
\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1-x^2}} = 0
we can rewrite it as
\frac{dy}{dx } =- \sqrt{\frac{1-y^2}{1-x^2}}\\ \\ \frac{dy}{\sqrt{1-y^2}}= \frac{-dx}{\sqrt{1-x^2}}
Now, integrate on both the sides
\sin^{-1}y + C =- \sin ^{-1}x + C'\\ \\ \sin^{-1}y+\sin^{-1}x= C
Therefore, the general solution of the differential equation \frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1-x^2}} = 0   is  \sin^{-1}y+\sin^{-1}x= C
 

Posted by

Gautam harsolia

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