Question

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

Answer 1

We have,

$\sqrt{x+y}+\sqrt{y-x}=c$ ……… (1)

On differentiating above equation w.r.t $x$, we get

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

$\frac{1}{2\sqrt{x+y}}+\frac{dy}{dx}\frac{1}{2\sqrt{x+y}}+\frac{1}{2\sqrt{y-x}}\frac{dy}{dx}-\frac{1}{2\sqrt{y-x}}=0$

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

$\left(\frac{\sqrt{y-x}+\sqrt{x+y}}{\sqrt{x+y}\sqrt{y-x}}\right)\frac{dy}{dx}=\left(\frac{\sqrt{x+y}-\sqrt{y-x}}{\sqrt{x+y}\sqrt{y-x}}\right)$

$\frac{dy}{dx}=\left(\frac{\sqrt{x+y}-\sqrt{y-x}}{\sqrt{x+y}+\sqrt{y-x}}\right)$

$\frac{dy}{dx}=\left(\frac{\sqrt{x+y}-\sqrt{y-x}}{\sqrt{x+y}+\sqrt{y-x}}\right)\times \left(\frac{\sqrt{x+y}-\sqrt{y-x}}{\sqrt{x+y}-\sqrt{y-x}}\right)$

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

$\frac{dy}{dx}=\frac{(x+y+y-x-2\sqrt{x+y}\times \sqrt{y-x})}{x+y-y+x}$

$\frac{dy}{dx}=\frac{(2y-2\sqrt{y+x}\times \sqrt{y-x})}{2x}$

$\frac{dy}{dx}=\frac{(y-\sqrt{y^2-x^2})}{x}$ ……… (2)

On differentiating above equation w.r.t $x$, we get

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

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

From equation (2), we get

$\frac{d^{2}y}{d{x}^2}=\frac{x(\left(\frac{(y-\sqrt{y^2-x^2})}{x}\right)-(\frac{1}{\sqrt{y^2-x^2}}\times (y\left(\frac{(y-\sqrt{y^2-x^2})}{x}\right)-x)))-(y-\sqrt{y^2-x^2})}{x^2}$

$\frac{d^{2}y}{d{x}^2}=\frac{((y-\sqrt{y^2-x^2})-(\frac{x}{\sqrt{y^2-x^2}}\times \left(\frac{y(y-\sqrt{y^2-x^2})-x^2}{x}\right)))-(y-\sqrt{y^2-x^2})}{x^2}$

$\frac{d^{2}y}{d{x}^2}=\frac{(y-\sqrt{y^2-x^2})-\left(\frac{y(y-\sqrt{y^2-x^2})-x^2}{\sqrt{y^2-x^2}}\right)-(y-\sqrt{y^2-x^2})}{x^2}$

$\frac{d^{2}y}{d{x}^2}=\frac{-\left(\frac{(y^2-y\sqrt{y^2-x^2})-x^2}{\sqrt{y^2-x^2}}\right)}{x^2}$

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

Hence, this is the answer.