Differential Equations Question 128
Question: The solution of differential equation $ \frac{dy}{dx}+y=1 $ is
[Pb. CET 2000]
Options:
A) $ y=1+c{e^{-x}} $
B) $ y=1-c{e^{-x}} $
C) $ y=x+c{e^{-x}} $
D) $ y=x-c{e^{-x}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{dy}{dx}+y=1 $ ; I.F. $ ={e^{\int{Pdx}}}={e^{\int{dx}}}=e^{x} $
Hence solution is $ y.e^{x}=\int{e^{x}dx+c} $
$ ye^{x}=e^{x}+c $
Therefore $ y=1+c{e^{-x}} $ .
BETA
