Question

Solution of differential equation $x^2=1+(\frac{x}{y}{)}^{-1}\frac{dy}{dx}+\frac{\left(\frac{x}{y}{)}^{-2}\right(\frac{dy}{dx}{)}^2}{2!}+\frac{\left(\frac{x}{y}{)}^{-3}\right(\frac{dy}{dx}{)}^3}{3!}+....$ is

Answer 1

Given,

$x^2=1+{\left(\frac{x}{y}\right)}^{-1}\frac{dy}{dx}+\frac{{\left(\frac{x}{y}\right)}^{-1}\frac{dy}{dx}}{2!}+\frac{{\left(\frac{x}{y}\right)}^{-1}\frac{dy}{dx}}{3!}+....$

$\Rightarrow x^2=1+{\left(\frac{x}{y}\right)}^{-1}\frac{dy}{dx}+\frac{{\left({\left(\frac{x}{y}\right)}^{-1}\frac{dy}{dx}\right)}^2}{2!}+\frac{{\left({\left(\frac{x}{y}\right)}^{-1}\frac{dy}{dx}\right)}^3}{3!}+....$

$\Rightarrow x^2=e^{{\left(\frac{x}{y}\right)}^{-1}\frac{dy}{dx}}$ $($Using $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....$ $)$

Taking log of both sides,

$\Rightarrow {\log }_e{x}^2={\left(\frac{x}{y}\right)}^{-1}\frac{dy}{dx}$

$\Rightarrow {\log }_e{x}^2.xdx=ydy$

Integrating both sides, we get,

$\Rightarrow \int {\log }_e{x}^2.xdx=\int ydy$

Putting $x^2=t$ on Left-hand side, we get,

$\Rightarrow \frac{1}{2}\int {\log }_{e}tdt=\int ydy$ $(Sincedt=2xdx)$

$\Rightarrow \frac{1}{2}t{\log }_{e}t-\frac{1}{2}t=\frac{y^2}{2}+C$ $(Using\int {\log }_{e}xdx=x{\log }_{e}x-x+C)$

$\Rightarrow \frac{1}{2}{x}^2{\log }_e{x}^2-\frac{1}{2}{x}^2=\frac{y^2}{2}+C$

$\Rightarrow y^2=2x^2{\log }_{e}x-x^2+C$

Hence the solution of given differential equation is $y^2=2x^2{\log }_{e}x-x^2+C$