Differential Equations Question 160
Question: The general solution of the differential equation $ (2x-y+1)dx+(2y-x+1)dy=0 $ is
[Karnataka CET 2005]
Options:
A) $ x^{2}+y^{2}+xy-x+y=c $
B) $ x^{2}+y^{2}-xy+x+y=c $
C) $ x^{2}-y^{2}+2xy-x+y=c $
D) $ x^{2}-y^{2}-2xy+x-y=c $
Show Answer
Answer:
Correct Answer: B
Solution:
$ (2x-y+1)dx+(2y-x+1)dy=0 $
$ \frac{dy}{dx}=\frac{2x-y+1}{x-2y-1} $ , put $ x=X+h $ , $ y=Y+k $
$ \frac{dY}{dX}=\frac{2X-Y+2h-k+1}{X-2Y+h-2k-1} $
$ 2h-k+1=0 $
Therefore $ h-2k-1=0 $
On solving $ h=-1 $ , $ k=-1 $ ; \ $ \frac{dY}{dX}=\frac{2X-Y}{X-2Y} $
Put $ Y=vX $ ; \ $ \frac{dY}{dX}=v+X\frac{dv}{dX} $
$ v+X\frac{dv}{dX}=\frac{2X-vX}{X-2vX}=\frac{2-v}{1-2v} $
$ X\frac{dv}{dX}=\frac{2-2v+2v^{2}}{1-2v}=\frac{2(v^{2}-v+1)}{1-2v} $
\ $ \frac{dX}{X}=\frac{(1-2v)}{2(v^{2}-v+1)}dv $
Put $ v^{2}-v+1=t $
Therefore $ (2v-1)dv=dt $
\ $ \frac{dX}{X}=-\frac{dt}{2t} $ \ $ \log X=\log {t^{-1/2}}+\log c $
\ $ X={t^{-1/2}}c $
Therefore $ X={{(v^{2}-v+1)}^{-1/2}}.c $
$ X^{2}(v^{2}-v+1)= $ constant $ {{(x+1)}^{2}}( \frac{{{(y+1)}^{2}}}{{{(x+1)}^{2}}}-\frac{(y+1)}{x+1}+1 )= $ constant $ {{(y+1)}^{2}}-(y+1)(x+1)+{{(x+1)}^{2}}=c $
$ y^{2}+x^{2}-xy+x+y=c $ .
BETA
