Differential Equations Question 155
Question: The solution of the equation $ \frac{dy}{dx}=\frac{x}{2y-x} $ is
Options:
A) $ (x-y){{(x+2y)}^{2}}=c $
B) $ y=x+c $
C) $ y=(2y-x)+c $
D) $ y=\frac{x}{2y-x}+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{dy}{dx}=\frac{x}{2y-x} $ . Put $ y=vx $
Therefore $ v+x\frac{dv}{dx}=\frac{dy}{dx} $
$ v+x\frac{dv}{dx}=\frac{x}{2v-x}=\frac{1}{2v-1} $
$ x\frac{dv}{dx}=\frac{1}{2v-1}-v=\frac{1-2v^{2}+v}{2v-1}=-\frac{(v-1)(2v+1)}{2v-1} $
$ \frac{(2v-1)}{(2v+1)(v-1)}=\frac{-dx}{x} $ ; $ \frac{1}{3(v-1)}+\frac{4}{3(2v+1)}=\frac{-dx}{x} $
$ \frac{1}{3}\log (v-1)+\frac{4}{3}.\frac{1}{2}\log (2v+1)=\log \frac{1}{x}+\log c $
$ \log {{(v-1)}^{1/3}}+\log {{(2v+1)}^{2/3}}=\log \frac{c}{x} $
$ ={{(v-1)}^{1/3}}{{(2v+1)}^{2/3}}=\frac{c}{x} $
$ ( \frac{y-x}{x} ){{( \frac{2y+x}{x} )}^{2}}=\frac{c^{3}}{x^{3}} $
Therefore $ (x-y){{(x+2y)}^{2}}=c $ .
BETA
