Differential Equations Question 149
Question: The general solution of the differential equation $ (x+y)dx+xdy=0 $ is
[MP PET 1994, 95]
Options:
A) $ x^{2}+y^{2}=c $
B) $ 2x^{2}-y^{2}=c $
C) $ x^{2}+2xy=c $
D) $ y^{2}+2xy=c $
Show Answer
Answer:
Correct Answer: C
Solution:
$ (x+y)dx+xdy=0 $
Therefore $ xdy=-(x+y)dx $
Therefore $ \frac{dy}{dx}=-\frac{x+y}{x} $
It is homogenous equation, hence put $ y=vx $ and $ \frac{dy}{dx}=v+x\frac{dv}{dx}, $ we get $ v+x\frac{dv}{dx}=-\frac{x+vx}{x}=-\frac{1+v}{1} $
Therefore $ x\frac{dv}{dx}=-1-2v $
Therefore $ \int _{{}}^{{}}{\frac{dv}{1+2v}}=-\int _{{}}^{{}}{\frac{dx}{x}} $
Therefore $ \frac{1}{2}\log (1+2v)=-\log x+\log c $
Therefore $ \log ( 1+2\frac{y}{x} )=2\log \frac{c}{x} $
Therefore $ \frac{x+2y}{x}={{( \frac{c}{x} )}^{2}} $
Therefore $ x^{2}+2xy=c $ .
BETA
