Here's the question you clicked on:
hary
du(x)/dx- tan(x)u(x)=-2sinx find u(x) inhomogeneous differential equation
\[u'-(\tan x)u=-2\sin x\\ (\cos x)u'-(\sin x)u=-2\sin x\cos x\] What can you say about the LHS?
I saw you were stuck on this one earlier hary, what was the book answer again?
u=cosx +C/cosx this was the book answer
Ah ok good. You must've made a small mistake near the end then. Because your integrating factor looked good.
hmm im guessing the same too..but couldnt figure out where
\[\large u \cos x=-2 \int\limits \sin x \cos x dx\] \[\large m=\cos x\]\[\large -dm=\sin x dx\] Applying this substitution gives us,\[\large u \cos x=2 \int\limits m dm\] \[\large u \cos x=m^2+C\] \[\large u \cos x=\cos^2x+C\]
owh god thats brilliant...did a blunder at my RHS integration
lol, what a mistake. Time for me to go to sleep I think
@zepdix thanks alot!!!! was cracking my head on this for quite long...pheww thats a relief... thankx again...
@myko thankx to you too...been a great help