Aspire Faculty ID #18335 · Topic: AMU MCA 2017 · Just now
AMU MCA 2017

If $y=\sqrt{\frac{1-x}{1+x}}$, then $(1-x^2)\frac{dy}{dx}+y$ is

Solution

$y^2=\frac{1-x}{1+x}$ Differentiate, $2y\frac{dy}{dx}=\frac{-(1+x)-(1-x)}{(1+x)^2}=\frac{-2}{(1+x)^2}$ $y\frac{dy}{dx}=\frac{-1}{(1+x)^2}$ Using $y^2=\frac{1-x}{1+x}$ Simplifying, $(1-x^2)\frac{dy}{dx}+y=0$

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