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

- anonymous

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

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- shubhamsrg

y= a cos2x ?

- shubhamsrg

guess that substn should work..

- anonymous

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

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## More answers

- shubhamsrg

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

- anonymous

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

- anonymous

heh.

- 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

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

- anonymous

yes, tat's it

- anonymous

that*

- anonymous

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

- anonymous

did you get it to work?

- anonymous

that's the particular soln. anyway...

- anonymous

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

- anonymous

why?

- anonymous

since sinx is twice? applying case 4 of undetermined coefficients

- anonymous

where it has some repeated imaginary root

- 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

this is for the particular solution...

- anonymous

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

- anonymous

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

- anonymous

lambda = 2i

- anonymous

ok?

- anonymous

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

- anonymous

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

- 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

ok:)

- anonymous

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

- anonymous

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

- anonymous

yes.

- anonymous

i forgot 2, lol.

- anonymous

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

- anonymous

yes. it would. just checked it.

- anonymous

thanks!

- anonymous

sure:)

- anonymous

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

- anonymous

just for the Yp part. swear.

- anonymous

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

- anonymous

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

- anonymous

hmm, I see your point ...

- anonymous

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

- anonymous

let me try it out and see...

- anonymous

sure. :)

- anonymous

seems to work... I got a= 2 b=-2

- anonymous

so 2cos^2 x -2sin^2 x

- anonymous

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

- anonymous

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

- anonymous

yes.

- anonymous

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

- anonymous

you left out and x I think...

- anonymous

i don't think so.

- anonymous

(1/2)*( sin^2 x - x*sin 2x ) ??

- anonymous

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

- anonymous

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

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