Question

If $y=\sqrt{\dfrac{1-x}{1+x}}$ , prove that $\left( 1-{{x}^{2}} \right)\dfrac{dy}{dx}+y=0$.

Answer 1

Hint: Problems like these can be solved by multiplying $\sqrt{\left( 1-x \right)}$ to both the numerator and denominator of the fraction in the right-hand side of given equation and further simplifying. After, getting the simplified expression of $y$ we differentiate both the sides of the equation using formula of differentiation $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}\left( u \right)-u\dfrac{d}{dx}\left( v \right)}{{{v}^{2}}}$ . Completing the differentiation and some simplifications we will reach to the desired differential equation.

Complete step-by-step answer:
The given expression we have is
$y=\sqrt{\dfrac{1-x}{1+x}}$
We further multiply $\sqrt{\left( 1-x \right)}$ to both the nominator and denominator of the fraction in the right-hand side of the above equation as shown below
$\Rightarrow y=\sqrt{\dfrac{\left( 1-x \right)\left( 1-x \right)}{\left( 1+x \right)\left( 1-x \right)}}$
$\Rightarrow y=\dfrac{1-x}{\sqrt{1-{{x}^{2}}}}.....\left( 1 \right)$
Now, while differentiating both the sides of the above equation we must use the formula for differentiating a function in the right-hand side
$\Rightarrow \dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}\left( u \right)-u\dfrac{d}{dx}\left( v \right)}{{{v}^{2}}}$


Applying this formula for differentiating equation $1$ we get
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\sqrt{1-{{x}^{2}}}\dfrac{d}{dx}\left( 1-x \right)-\left( 1-x \right)\dfrac{d}{dx}\left( \sqrt{1-{{x}^{2}}} \right)}{1-{{x}^{2}}}$
Further, doing all the differentiations we get
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\sqrt{1-{{x}^{2}}}\left( -1 \right)-\left( 1-x \right)\left\{ \dfrac{1}{2}{{\left( 1-{{x}^{2}} \right)}^{-\dfrac{1}{2}}}\left( -2x \right) \right\}}{1-{{x}^{2}}}$

Simplifying the right-hand side of the above equation we get
$\Rightarrow \dfrac{dy}{dx}=\dfrac{x\left( 1-x \right)-\left( 1-{{x}^{2}} \right)}{\left( 1-{{x}^{2}} \right)\sqrt{1-{{x}^{2}}}}$
Omitting the brackets of the numerator of right-hand side of the above equation we get
$\Rightarrow \dfrac{dy}{dx}=\dfrac{x-{{x}^{2}}-1+{{x}^{2}}}{\left( 1-{{x}^{2}} \right)\sqrt{1-{{x}^{2}}}}$
Summing up the like terms the equation becomes
$\Rightarrow \dfrac{dy}{dx}=\dfrac{x-1}{\left( 1-{{x}^{2}} \right)\sqrt{1-{{x}^{2}}}}$
Taking $-1$ common from the numerator of the right-hand side of the above equation
$\Rightarrow \dfrac{dy}{dx}=-\dfrac{1-x}{\left( 1-{{x}^{2}} \right)\sqrt{1-{{x}^{2}}}}$
Multiplying both the right and left-hand side of the above equation with $\sqrt{1-{{x}^{2}}}$ we get

 $\Rightarrow \left( 1-{{x}^{2}} \right)\dfrac{dy}{dx}=-\dfrac{1-x}{\sqrt{1-{{x}^{2}}}}$
Now, substituting the value of $y$ in the above expression in place of $\dfrac{1-x}{\sqrt{1-{{x}^{2}}}}$ we get
$\Rightarrow \left( 1-{{x}^{2}} \right)\dfrac{dy}{dx}=-y$
Adding $y$ to both the sides of the above equation we get
$\Rightarrow \left( 1-{{x}^{2}} \right)\dfrac{dy}{dx}+y=-y+y$
$\Rightarrow \left( 1-{{x}^{2}} \right)\dfrac{dy}{dx}+y=0$
Thus, we can prove $\left( 1-{{x}^{2}} \right)\dfrac{dy}{dx}+y=0$ , if $y=\sqrt{\dfrac{1-x}{1+x}}$ .

Note: As the given problem contains too complex an expression having square roots and fractions, we should be very careful while differentiating them. Most of the time, squaring on both sides makes the solution easier and helps us to avoid complex forms. The problem can also be solved by taking natural logarithm on both sides and then differentiating on both sides.