Differential Equations Question 223
Question: If $ \phi (x) $ is a differentiable function, then the solution of the differential equation $ dy+{y\phi ‘(x)-\phi (x)\phi ‘(x)}dx=0 $ is
Options:
A) $ y={\phi (x)-1}+c{e^{-\phi (x)}} $
B) $ y\phi (x)={{{\phi (x)}}^{2}}+c $
C) $ y{e^{\phi (x)}}=\phi (x){e^{\phi (x)}}+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] We have, $ dy+{y\phi ‘(x)-\phi (x)\phi ‘(x)}dx=0 $
$ \Rightarrow \frac{dy}{dx}+\phi ‘(x)\cdot y=\phi (x)\phi ‘(x) $
- (i) This is linear differential equation with I.F $ ={e^{\int{\phi ‘(x)dx}}}={e^{\phi (x)}} $ Multiplying Eq. (i) by $ \phi (x) $ and integrating, we get $ y{e^{\phi (x)}}=\int{\phi (x)\phi ‘(x){e^{\phi (x)}}}dx $
$ \Rightarrow y{e^{\phi (x)}}=\int{{e^{\phi (x)}}\phi (x)\phi ‘(x)}dx $
$ \Rightarrow y{e^{\phi (x)}}=\int{\phi (x){e^{\phi (x)}}}\phi ‘(x)dx $
$ \Rightarrow y{e^{\phi (x)}}=\phi (x){e^{\phi (x)}}-\int{\phi ‘(x){e^{\phi (x)}}}dx $
$ \Rightarrow y{e^{\phi (x)}}=\phi (x){e^{\phi (x)}}-{e^{\phi (x)}}+c $
$ \Rightarrow y={\phi (x)-1}+c{e^{-\phi (x)}} $
BETA
