Question

Verify that $y^2=4a(x+a)$ is a solution of the differential equation $y\left\{1-{\left(\frac{dy}{dx}\right)}^2\right\}=2x\frac{dy}{dx}$.

Answer 1

Step $1:$ Differentiation:

Differentiating $y^2=4a(x+a)$ w.r.t. x:

$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{y}}^2\right)& =& \frac{\mathrm{d}}{\mathrm{dx}}\left(4\mathrm{a}(\mathrm{x}+\mathrm{a})\right)\\ 2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}& =& \frac{\mathrm{d}}{\mathrm{dx}}\left(4\mathrm{ax}\right)+\frac{\mathrm{d}}{\mathrm{dx}}\left(4{\mathrm{a}}^2\right)\\ 2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}& =& 4\mathrm{a}+0\\ \frac{\mathrm{dy}}{\mathrm{dx}}& =& \frac{2\mathrm{a}}{\mathrm{y}}\end{array}$

Step $2:$ Verifying $y^2=4a(x+a)$ is a solution of the differential equation $y\left\{1-{\left(\frac{dy}{dx}\right)}^2\right\}=2x\frac{dy}{dx}$:

Taking L.H.S., we get,

$\begin{array}{rcl}\mathrm{y}\left\{1-{\left(\frac{d\mathrm{y}}{d\mathrm{x}}\right)}^2\right\}& =& \mathrm{y}\left\{1-{\left(\frac{2\mathrm{a}}{\mathrm{y}}\right)}^2\right\}\\ & =& \mathrm{y}\left\{1-\left(\frac{4{\mathrm{a}}^2}{{\mathrm{y}}^2}\right)\right\}\\ & =& \mathrm{y}\left\{\frac{{\mathrm{y}}^2-4{\mathrm{a}}^2}{{\mathrm{y}}^2}\right\}\\ & =& \left\{\frac{4\mathrm{a}(\mathrm{x}+\mathrm{a})-4{\mathrm{a}}^2}{\mathrm{y}}\right\}\left(\because y^2=4a(x+a)\right)\\ & =& \left\{\frac{4\mathrm{ax}+4{\mathrm{a}}^2-4{\mathrm{a}}^2}{\mathrm{y}}\right\}\\ & =& \left\{\frac{4\mathrm{ax}}{\mathrm{y}}\right\}\\ & =& 2\mathrm{x}\left\{\frac{2\mathrm{a}}{\mathrm{y}}\right\}\\ & =& 2\mathrm{x}\frac{d\mathrm{y}}{d\mathrm{x}}\\ & =& \mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Hence, $y^2=4a(x+a)$ is a solution of the differential equation $y\left\{1-{\left(\frac{dy}{dx}\right)}^2\right\}=2x\frac{dy}{dx}$.