Differential Equations Question 121
Question: The differential equation found by the elimination of the arbitrary constant K from the equation $ y=(x+K){e^{-x}} $ is
Options:
A) $ \frac{dy}{dx}-y={e^{-x}} $
B) $ \frac{dy}{dx}-ye^{x}=1 $
C) $ \frac{dy}{dx}+ye^{x}=1 $
D) $ \frac{dy}{dx}+y={e^{-x}} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ y=(x+K){e^{-x}} $
Therefore $ \frac{dy}{dx}=-(x+K){e^{-x}}+{e^{-x}} $
Therefore $ \frac{dy}{dx}=-y+{e^{-x}} $
Therefore $ \frac{dy}{dx}+y={e^{-x}} $ .
BETA
