Differential Equations Question 50
Question: Solution of $ (x+y-1)dx+(2x+2y-3)dy=0 $ is
[MP PET 1999]
Options:
A) $ y+x+\log (x+y-2)=c $
B) $ y+2x+\log (x+y-2)=c $
C) $ 2y+x+\log (x+y-2)=c $
D) $ 2y+2x+\log (x+y-2)=c $
Show Answer
Answer:
Correct Answer: C
Solution:
Given equation is $ \frac{dy}{dx}=-( \frac{x+y-1}{2x+2y-3} ) $
Put $ x+y=t $
Therefore $ \frac{dy}{dx}=\frac{dt}{dx}-1 $
$ \therefore \frac{dy}{dx}=\frac{1-t}{2t-3} $
Therefore $ \frac{dt}{dx}-1=\frac{1-t}{2t-3} $
Therefore $ \frac{dt}{dx}=\frac{t-2}{2t-3} $
Therefore $ \frac{2t-3}{t-2}dt=dx $ . Integrating both sides, we get $ \int _{{}}^{{}}{\frac{2t-4}{t-2}dt}-\int _{{}}^{{}}{\frac{3-4}{t-2}dt}=\int _{{}}^{{}}{1}dx $
Therefore $ 2t+\log (t-2)=x+c $
Therefore $ 2(x+y)+\log (x+y-2)=x+c $
Therefore $ 2y+x+\log (x+y-2)=c $ .
BETA
