Differential Equations Question 73
Question: The solution of $ \frac{dy}{dx}+y={e^{-x}},y(0)=0 $ , is
[Kerala (Engg.) 2002]
Options:
A) $ y={e^{-x}}(x-1) $
B) $ y=xe^{x} $
C) $ y=x{e^{-x}}+1 $
D) $ y=x{e^{-x}} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{dy}{dx}+y={e^{-x}} $ ; I.F. $ ={e^{\int{dx}}}=e^{x} $
\ $ ye^{x}=\int{{e^{-x}}.e^{x}dx+c} $
Therefore $ ye^{x}=x+c $
Since $ y(0)=0 $ , \ $ c=0 $
Hence, the required solution is $ ye^{x}=x $
Therefore $ y=x{e^{-x}} $ .
BETA
