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



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shubhamsrg
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y= a cos2x ?

shubhamsrg
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guess that substn should work..

lilMissMindset
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aaa. i think i got it. wait, i'll try.

shubhamsrg
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ohh wait,,i didnt see a "y" also in there. !!

Algebraic!
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@shubhamsrg
I think it's just written in 'operator' notation...
\[\frac{ d^2y }{dt^2 } +4y = 4\sin^2x\]
correct @lilMissMindset ?

Algebraic!
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heh.

shubhamsrg
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there's a y also with D^2 ..
i had got that,,i didnt see a y being multiplied to whole that time..sorry..

Algebraic!
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\[\frac{ d^2y }{ dx^2 } +4y =4\sin^2x\]
rather.

lilMissMindset
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yes, tat's it

lilMissMindset
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that*

Algebraic!
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yes he's correct I believe y =a*cos^2 x

Algebraic!
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did you get it to work?

Algebraic!
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that's the particular soln. anyway...

lilMissMindset
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isn't it like, Acosx+Bsinx+Cxcosx+Dxsinx

Algebraic!
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why?

lilMissMindset
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since sinx is twice? applying case 4 of undetermined coefficients

lilMissMindset
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where it has some repeated imaginary root

Algebraic!
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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)

Algebraic!
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this is for the particular solution...

Algebraic!
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finding the homogeneous solution shouldn't be a problem...

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\[\lambda ^2 +4 =0\]

Algebraic!
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lambda = 2i

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ok?

lilMissMindset
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i'll work on it. please guide me still, in case i get stuck.

Algebraic!
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hold on... can you tell me what part is giving you problems?

lilMissMindset
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getting Yp gives me a lot of trouble. since you told me that it's acos^2 x, i am differentiating it now.

Algebraic!
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ok:)

lilMissMindset
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how about if the given is cos^x, then Yp=asin^2 x?

Algebraic!
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if the forcing term is cos^2 x you mean?

lilMissMindset
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yes.

lilMissMindset
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i forgot 2, lol.

Algebraic!
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yeah, for the same equation, I believe that a*sin^2 x would be the trial solution then...

Algebraic!
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yes. it would. just checked it.

lilMissMindset
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thanks!

Algebraic!
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sure:)

lilMissMindset
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darn, i can't do it. show me a brief solution please. so i can learn how.

lilMissMindset
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just for the Yp part. swear.

Algebraic!
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well, I did it already
(a*cos^2(x))" +4a cos^2(x) = 4sin^2(x)

Algebraic!
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2a*cos^2(x)  2a*sin^2(x) +4a cos^2(x) = 4sin^2(x)

Algebraic!
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hmm, I see your point ...

Algebraic!
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I guess you'll be needing something like yp = a*cos^2(x) + b*sin^2(x)

Algebraic!
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let me try it out and see...

lilMissMindset
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sure. :)

Algebraic!
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seems to work... I got a= 2 b=2

Algebraic!
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so 2cos^2 x 2sin^2 x

lilMissMindset
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answer here in the book is
1/2(1xsin2x) for the Yp part.

Algebraic!
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I'll assume that's (1/2)*( 1  x*sin^2 x )

lilMissMindset
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yes.

Algebraic!
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you should check the original, I think you might have written the problem wrong...

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you left out and x I think...

lilMissMindset
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i don't think so.

Algebraic!
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(1/2)*( sin^2 x  x*sin 2x ) ??

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I don't see how
(1/2)(1xsin2x)
can work... maybe I'm missing something.

Algebraic!
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something's off in your original equation or that soln.