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DIFFERENTIAL EQUATION: help please, anybody. (D^2 +4)y =4sin^2 (x)

Mathematics
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y= a cos2x ?
guess that substn should work..
aaa. i think i got it. wait, i'll try.

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Other answers:

ohh wait,,i didnt see a "y" also in there. !!
@shubhamsrg I think it's just written in 'operator' notation... \[\frac{ d^2y }{dt^2 } +4y = 4\sin^2x\] correct @lilMissMindset ?
heh.
there's a y also with D^2 .. i had got that,,i didnt see a y being multiplied to whole that time..sorry..
\[\frac{ d^2y }{ dx^2 } +4y =4\sin^2x\] rather.
yes, tat's it
that*
yes he's correct I believe y =a*cos^2 x
did you get it to work?
that's the particular soln. anyway...
isn't it like, Acosx+Bsinx+Cxcosx+Dxsinx
why?
since sinx is twice? applying case 4 of undetermined coefficients
where it has some repeated imaginary root
look d^2 (a*cos^2(x)) /dx^2 +4a cos^2(x) = 4sin^2(x) second derivative of a*cos^2(x) is 2a*cos^2(x) - 2a*sin^2(x)
this is for the particular solution...
finding the homogeneous solution shouldn't be a problem...
\[\lambda ^2 +4 =0\]
lambda = 2i
ok?
i'll work on it. please guide me still, in case i get stuck.
hold on... can you tell me what part is giving you problems?
getting Yp gives me a lot of trouble. since you told me that it's acos^2 x, i am differentiating it now.
ok:)
how about if the given is cos^x, then Yp=asin^2 x?
if the forcing term is cos^2 x you mean?
yes.
i forgot 2, lol.
yeah, for the same equation, I believe that a*sin^2 x would be the trial solution then...
yes. it would. just checked it.
thanks!
sure:)
darn, i can't do it. show me a brief solution please. so i can learn how.
just for the Yp part. swear.
well, I did it already (a*cos^2(x))" +4a cos^2(x) = 4sin^2(x)
2a*cos^2(x) - 2a*sin^2(x) +4a cos^2(x) = 4sin^2(x)
hmm, I see your point ...
I guess you'll be needing something like yp = a*cos^2(x) + b*sin^2(x)
let me try it out and see...
sure. :)
seems to work... I got a= 2 b=-2
so 2cos^2 x -2sin^2 x
answer here in the book is 1/2(1-xsin2x) for the Yp part.
I'll assume that's (1/2)*( 1 - x*sin^2 x )
yes.
you should check the original, I think you might have written the problem wrong...
you left out and x I think...
i don't think so.
(1/2)*( sin^2 x - x*sin 2x ) ??
I don't see how (1/2)(1-xsin2x) can work... maybe I'm missing something.
something's off in your original equation or that soln.

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