Question

If $\sqrt{x+y}-\sqrt{x-y}=c$, then $\frac{d^{2}y}{d{x}^2}=$

• A. $\frac{1}{y}{\left(\frac{dy}{dx}\right)}^2$
• B. $\frac{-c^4}{4y^3}$
• C. $y{\left(\frac{dy}{dx}\right)}^2$
• D. $\frac{-c^2}{4y^3}$

Answer 1

Answer: (B)

It is given that
$\sqrt{x+y}-\sqrt{x-y}=c$
On squaring both sides, we get
$(x+y)+(x-y)-2\sqrt{x^2-y^2}=c^2$
$\Rightarrow 2x-c^2=2\sqrt{x^2-y^2}$
Again on squaring both sides, we get
$4x^2+c^4-4x{c}^2=4x^2-4y^2$
$\Rightarrow c^4-4x{c}^2=-4y^2$
$\Rightarrow 4y^2=4x{c}^2-c^4$
On differentiating w.r.t. $x$, we get
$8y\frac{dy}{dx}=4c^2$
$\Rightarrow 2y\frac{dy}{dx}=c^2...(i)$
On differentiating both sides w.r.t. $x$, we get
$2{\left(\frac{dy}{dx}\right)}^2+2y\frac{d^{2}y}{d{x}^2}=0$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=-\frac{{\left(\frac{dy}{dx}\right)}^2}{y}=-\frac{{\left(\frac{c^2}{2y}\right)}^2}{y}[$ From Eq.(i)]
$=-\frac{c^4}{4y^3}$