Differential Equations Question 66
Question: The solution of $ \log ( \frac{dy}{dx} )=ax+by $ is
[AMU 2002]
Options:
A) $ \frac{e^{by}}{b}=\frac{e^{ax}}{a}+c $
B) $ \frac{{e^{-by}}}{-b}=\frac{e^{ax}}{a}+c $
C) $ \frac{{e^{-by}}}{a}=\frac{e^{ax}}{b}+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \log ( \frac{dy}{dx} )=ax+by $
Therefore $ \frac{dy}{dx}={e^{ax+by}}=e^{ax}.e^{by} $
Therefore $ {e^{-by}}dy=e^{ax}dx $
Therefore $ \frac{{e^{-by}}}{-b}=\frac{e^{ax}}{a}+c $ .
BETA
