Differential Equations Question 168
Question: Solution of differential equation $ x^{2}=1+{{( \frac{x}{y} )}^{-1}}\frac{dy}{dx}+\frac{{{( \frac{x}{y} )}^{-2}}{{( \frac{dy}{dx} )}^{2}}}{2!} $ $ +\frac{{{( \frac{x}{y} )}^{-3}}{{( \frac{dy}{dx} )}^{3}}}{3!}+……… $ is
Options:
A) $ y^{2}=x^{2}(lnx^{2}-1)+C $
B) $ y=x^{2}(lnx-1)+C $
C) $ y^{2}=x(lnx-1)+C $
D) $ y=x^{2}{e^{x^{2}}}+C $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ x^{2}={e^{{{( \frac{x}{y} )}^{-1}}( \frac{dy}{dx} )}}\Rightarrow x^{2}={e^{( \frac{y}{x} )( \frac{dy}{dx} )}} $
$ \Rightarrow $ ln $ x^{2}=\frac{y}{x}\frac{dy}{dx} $ or $ \int{xlnx^{2}dx=\int{ydy}} $ Put $ x^{2}=t\Rightarrow 2xdx=dt\therefore \frac{1}{2}\int{lntdt=\frac{y^{2}}{2}} $
$ C+t\ln ort-t=y^{2}ory^{2}=x^{2}(lnx^{2}-1)+C $
BETA
