anonymous
  • anonymous
DIFFERENTIAL EQUATION: help please, anybody. (D^2 +4)y =4sin^2 (x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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shubhamsrg
  • shubhamsrg
y= a cos2x ?
shubhamsrg
  • shubhamsrg
guess that substn should work..
anonymous
  • anonymous
aaa. i think i got it. wait, i'll try.

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

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