Differential Equations Question 242
Question: The solution of the differential equation $ x\sin x\frac{dy}{dx}+(x\cos x+\sin x)y=\sin x $ . When $ y(0)=0 $ is
Options:
A) $ xy\sin x=1-\cos x $
B) $ xy\sin x+\cos x=0 $
C) $ x\sin x+y\cos x=0 $
D) $ x\sin x+y\cos x=1 $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] The equation is $ \frac{dy}{dx}+( \frac{x\cos x+\sin x}{x\sin x} )y=\frac{1}{x} $ Integrating factor I.F. $ ={e^{\int{\frac{x\cos x+\sin x}{x\sin x}dx}}}={e^{\log (xsinx)}}=x\sin x $
$ \therefore $ The solution is $ y(xsinx)=\int{\frac{1}{x}(xsinx)dx+c} $
$ xy\sin x=-\cos x+c $ when $ x=0,y=0\Rightarrow c=\cos 0=1 $
$ \therefore $ The particular solution is $ xy\sin x=1-\cos x $
BETA
