Differential Equations Question 16
Question: The solution of the differential equation $ \frac{dy}{dx}+y=\cos x $ is
[AISSE 1990]
Options:
A) $ y=\frac{1}{2}(\cos x+\sin x)+c{e^{-x}} $
B) $ y=\frac{1}{2}(\cos x-\sin x)+c{e^{-x}} $
C) $ y=\cos x+\sin x+c{e^{-x}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
It is linear equation of the form $ \frac{dy}{dx}+Py=Q $
So, I.F. $ ={e^{\int _{{}}^{{}}{1dx}}}=e^{x} $
Hence solution is $ y.e^{x}=\int _{{}}^{{}}{\cos x.e^{x}dx+c} $
Therefore $ y=\frac{1}{2}(\cos x+\sin x)+c{e^{-x}} $ .
BETA
