Question

State true or false.

If $y=\sqrt{\frac{1-x}{1+x}}\cdot$ then $(1-x^2)\frac{dy}{dx}+y=0$

• A. True
• B. False

Answer 1

The correct option is B False

$y=\sqrt{\frac{1-x}{1+x}}$

$\Rightarrow y=\frac{\sqrt{1-x}}{\sqrt{1+x}}\times \frac{\sqrt{1-x}}{\sqrt{1-x}}$

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

Squaring both sides, we get

$\Rightarrow (1-x^2)y^2={(1-x)}^2$

Differentiating w.r.t $x$ both sides,

$\Rightarrow (1-x^2)2y\frac{dy}{dx}+y^2(-2x)=-2(1-x)$

Dividing both sides by $2$ we get

$\Rightarrow (1-x^2)y\frac{dy}{dx}-x{y}^2+(1-x)=0$

Hence the given statement is false.