Differential Equations Question 302
Question: The general solution of $ (x+1)\frac{dy}{dx}+1=2{e^{-y}} $ is
Options:
A) $ e^{y}(x+1)=x+C $
B) $ {e^{-y}}=2x+C $
C) $ e^{y}(x+1)=2x+C $
D) $ e^{y}(x+1)=C $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ (x+1)\frac{dy}{dx}+1=2{e^{-y}} $
$ \Rightarrow e^{y}\frac{dy}{dx}+\frac{e^{y}}{x+1}=\frac{2}{x+1} $ Put $ e^{y}=u\Rightarrow e^{y}\frac{dy}{dx}=\frac{du}{dx}\therefore \frac{du}{dx}+\frac{u}{x+1}=\frac{2}{x+1} $ I.F. $ ={e^{\int{\frac{1}{x+1}dx}}}={e^{\log (x+1)}}=x+1 $
$ \therefore $ Solution is $ u(x+1)=\int{\frac{2}{(x+1)}(x+1)dx+C} $
$ \Rightarrow e^{y}(x+1)=2x+C $
BETA
