Question
Solution of differential equation $${x^2} = 1 + {\left( {\frac{x}{y}} \right)^{ - 1}}\frac{{dy}}{{dx}} + \frac{{{{\left( {\frac{x}{y}} \right)}^{ - 2}}{{\left( {\frac{{dy}}{{dx}}} \right)}^2}}}{{2!}} + \frac{{{{\left( {\frac{x}{y}} \right)}^{ - 3}}{{\left( {\frac{{dy}}{{dx}}} \right)}^3}}}{{3!}} + .....\,{\text{is}}\,{\text{:}}$$
A.
$${y^2} = {x^2}\left( {\ln {x^2} - 1} \right) + C$$
B.
$$y = {x^2}\left( {\ln x - 1} \right) + C$$
C.
$${y^2} = x\left( {\ln x - 1} \right) + C$$
D.
$$y = {x^2}{e^{{x^2}}} + C$$
Answer :
$${y^2} = {x^2}\left( {\ln {x^2} - 1} \right) + C$$
Solution :
$$\eqalign{
& {x^2} = {e^{{{\left( {\frac{x}{y}} \right)}^{ - 1}}\left( {\frac{{dy}}{{dx}}} \right)}} \Rightarrow {x^2} = {e^{\left( {\frac{y}{x}} \right)\left( {\frac{{dy}}{{dx}}} \right)}} \cr
& \Rightarrow \ln \,{x^2} = \frac{y}{x}\frac{{dy}}{{dx}}{\text{ or }}\int {x\,\ln \,{x^2}dx} = \int {y\,dy} \cr
& {\text{Put }}{x^2} = t\, \Rightarrow 2x\,dx = dt \cr
& \therefore \frac{1}{2}\int {\ln \,t\,dt = \frac{{{y^2}}}{2}} \cr
& C + t\,\ln \,t - t = {y^2} \cr
& {\text{or }}{y^2} = {x^2}\left( {\ln {x^2} - 1} \right) + C \cr} $$