Differential Equations Question 158
Question: Solution of differential equation $ \frac{dy}{dx}=\frac{y-x}{y+x} $ is
[MP PET 1997]
Options:
A) $ {\log _{e}}(x^{2}+y^{2})+2{{\tan }^{-1}}\frac{y}{x}+c=0 $
B) $ \frac{y^{2}}{2}+xy=xy-\frac{x^{2}}{2}+c $
C) $ ( 1+\frac{x}{y} )y=( 1-\frac{x}{y} )x+c $
D) $ y=x-2{\log _{e}}y+c $
Show Answer
Answer:
Correct Answer: A
Solution:
Put $ y=vx $
Therefore $ \frac{dy}{dx}=v+x\frac{dv}{dx} $
$ \therefore v+x\frac{dv}{dx}=\frac{v-1}{v+1} $
Therefore $ x\frac{dv}{dx}=\frac{v-1}{v+1}-v $
Therefore $ x\frac{dv}{dx}=-\frac{v^{2}+1}{v+1} $
Therefore $ \int _{{}}^{{}}{\frac{dx}{x}}=-\int _{{}}^{{}}{\frac{v+1}{v^{2}+1}}dv $
Therefore $ -{\log _{e}}x=\frac{1}{2}\int _{{}}^{{}}{\frac{2v}{v^{2}+1}}dv+\int _{{}}^{{}}{\frac{1}{v^{2}+1}}dv $
Therefore $ -{\log _{e}}x=\frac{1}{2}\log (v^{2}+1)+{{\tan }^{-1}}v+c $
Therefore $ -2{\log _{e}}x=\log ( \frac{x^{2}+y^{2}}{x^{2}} )+2{{\tan }^{-1}}( \frac{y}{x} )+c $
Therefore $ {\log _{e}}(x^{2}+y^{2})+2{{\tan }^{-1}}\frac{y}{x}+c=0 $ .
BETA
