Differential Equations Question 147
Question: The solution of the differential equation $ (3xy+y^{2})dx+(x^{2}+xy)dy=0 $ is
[AISSE 1990]
Options:
A) $ x^{2}(2xy+y^{2})=c^{2} $
B) $ x^{2}(2xy-y^{2})=c^{2} $
C) $ x^{2}(y^{2}-2xy)=c^{2} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
It can be written in the form of homogeneous equation $ \frac{dy}{dx}=-\frac{3xy+y^{2}}{x^{2}+xy} $
So, now put $ y=vx $ and $ \frac{dy}{dx}=v+x\frac{dv}{dx}, $ we get $ v+x\frac{dv}{dx}=-\frac{3x^{2}v+x^{2}v^{2}}{x^{2}+x^{2}v} $
Therefore $ x\frac{dv}{dx}=\frac{-2v(v+2)}{v+1} $
Therefore $ \frac{1}{x}dx=-\frac{v+1}{2v(v+2)}dv=-[ \frac{1}{2(v+2)}+\frac{1}{2v(v+2)} ]dv $
Therefore $ -\frac{2}{x}dx=[ \frac{1}{v+2}+\frac{1}{2v}-\frac{1}{2v(v+2)} ] $
On integrating, we get $ -2{\log _{e}}x=\frac{1}{2}\log (v+2)+\frac{1}{2}\log v+\log c $
Therefore $ v(v+2)x^{4}=c^{2} $
Therefore $ \frac{y}{x}( \frac{y}{x}+2 )x^{4}=c^{2} $ , $ ( \because v=\frac{y}{x} ) $
Hence required solution is $ (y^{2}+2xy)x^{2}=c^{2} $ .
BETA
