Question

Given that $y=\sqrt{\frac{1-x}{1+x}}$ show that $(1-x^2)\frac{dy}{dx}+y=0$

Answer 1

Differentiating with respect to x,

$\begin{array}{c}y=\sqrt{\frac{1-x}{1+x}}\\ Let,\\ x=\sin \theta \\ So,\\ y=\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}\\ =\sqrt{\frac{1-\sin \theta }{1+\sin \theta }\times \frac{1-\sin \theta }{1-\sin \theta }}\\ =\sqrt{\frac{{\left(1-\sin \theta \right)}^2}{1^2-{\sin }^2\theta }}\\ =\sqrt{\frac{{\left(1-\sin \theta \right)}^2}{{\cos }^2\theta }}\\ =\frac{1-\sin \theta }{\cos \theta }\\ =\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta }\\ =\sec \theta -\tan \theta \\ \frac{dy}{dx}=\left(\sec \theta \tan \theta -{\sec }^2\theta \right)\times \frac{d\theta }{dx}\\ =\sec \theta \left(\tan \theta -\sec \theta \right)\times \frac{1}{\frac{d}{d\theta }\left(\sin \theta \right)}\\ =\sec \theta \left(-y\right)\times \frac{1}{\cos \theta }\\ =\frac{-y}{{\cos }^2\theta }\\ =\frac{-y}{1-{\sin }^2\theta }\\ \frac{dy}{dx}=\frac{-y}{1-x^2}\\ \left(1-x^2\right)\frac{dy}{dx}+y=0\end{array}$

Which is equal to the right hand side.

Hence Proved