Question

If y =root of 1-x/1+x then prove that (1-x^2)dy/dx+y=0

Answer 1

Please see the attachment

Answer 2

Answer:

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

Step-by-step explanation:

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

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

Proof:

The given equation is

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

Squaring both sides we get

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

Differential with respect to x.

$2y\frac{dy}{dx}=\frac{(1+x)\frac{d}{dx}(1-x)-(1-x)\frac{d}{dx}(1+x)}{(1+x)^2}$           $[\because (\frac{f}{g})'=\frac{g\cdot f'-f\cdot g'}{g^2}]$

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

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

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

Divide both sides by 2y.

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

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

We need to prove

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

Taking LHS,

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

Substitute the value of $\frac{dy}{dx}$.

$LHS=(1-x^2)(-\frac{1}{y(1+x)^2})+y$

$LHS=-\frac{(1-x)(1+x)}{y(1+x)^2}+y$

Cancel out the common factors.

$LHS=-\frac{1-x}{y(1+x)}+y$

$LHS=-\frac{1}{y}\times \frac{1-x}{(1+x)}+y$

$LHS=-\frac{1}{y}\times y^2+y$           $[\because y^2=\frac{1-x}{(1+x)}]$

Cancel out the common factors.

$LHS=-y+y$

$LHS=0$

$LHS=RHS$

Hence proved.