Question

If $\sqrt{y}-\frac{1}{\sqrt{y}}=2x,$ then which of the following options is correct ?

• A. $(x^2+1)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-4y=0$
• B. $(x^2-1)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-4y=0$
• C. $(x^2+1)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+4y=0$
• D. $(x^2+1)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-4y=0$

Answer 1

The correct option is A $(x^2+1)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-4y=0$

$\sqrt{y}-\frac{1}{\sqrt{y}}=2x$
$\Rightarrow (\sqrt{y}{)}^2-2x\sqrt{y}-1=0$
$\Rightarrow \sqrt{y}=\frac{2x+\sqrt{4x^2+4}}{2}(\because \sqrt{y}>0)$
$\Rightarrow y=(x+\sqrt{x^2+1}{)}^2$
$\Rightarrow \frac{dy}{dx}=2(x+\sqrt{x^2+1})(1+\frac{x}{\sqrt{x^2+1}})$
$\Rightarrow \frac{dy}{dx}=\frac{2(x+\sqrt{x^2+1}{)}^2}{\sqrt{x^2+1}}$
$\Rightarrow \frac{dy}{dx}=\frac{2y}{\sqrt{x^2+1}}$

Squaring both the sides
$(x^2+1){\left(\frac{dy}{dx}\right)}^2=4y^2$

Differentiate w.r.t. $x$
$(x^2+1)\times 2\left(\frac{dy}{dx}\right)\left(\frac{d^{2}y}{d{x}^2}\right)+2x{\left(\frac{dy}{dx}\right)}^2=8y\frac{dy}{dx}$

Dividing by $2\frac{dy}{dx}$, we get
$(x^2+1)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=4y$
$\Rightarrow (x^2+1)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-4y=0$