Differential Equations Question 333
Question: The solution of the differential equation $ \frac{dy}{dx}=1+x+y+xy $ is
[AISSE 1985; AI CBSE 1990; MP PET 2003]
Options:
A) $ \log (1+y)=x+\frac{x^{2}}{2}+c $
B) $ {{(1+y)}^{2}}=x+\frac{x^{2}}{2}+c $
C) $ \log (1+y)=\log (1+x)+c $
D) None of these
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Answer:
Correct Answer: A
Solution:
$ \frac{dy}{dx}=1+x+y+xy $
Therefore $ \frac{dy}{dx}=(1+x)(1+y) $
Therefore $ \frac{dy}{dx}+\sin ( \frac{x+y}{2} )=\sin ( \frac{x-y}{2} ) $
On integrating, we get $ \log (1+y)=\frac{x^{2}}{2}+x+c $ .
BETA
