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

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shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0guess that substn should work..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0aaa. i think i got it. wait, i'll try.

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0ohh wait,,i didnt see a "y" also in there. !!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@shubhamsrg I think it's just written in 'operator' notation... \[\frac{ d^2y }{dt^2 } +4y = 4\sin^2x\] correct @lilMissMindset ?

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0there's a y also with D^2 .. i had got that,,i didnt see a y being multiplied to whole that time..sorry..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ d^2y }{ dx^2 } +4y =4\sin^2x\] rather.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes he's correct I believe y =a*cos^2 x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0did you get it to work?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's the particular soln. anyway...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0isn't it like, Acosx+Bsinx+Cxcosx+Dxsinx

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0since sinx is twice? applying case 4 of undetermined coefficients

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0where it has some repeated imaginary root

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0look 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
 4 years ago
Best ResponseYou've already chosen the best response.0this is for the particular solution...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0finding the homogeneous solution shouldn't be a problem...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i'll work on it. please guide me still, in case i get stuck.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hold on... can you tell me what part is giving you problems?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0getting Yp gives me a lot of trouble. since you told me that it's acos^2 x, i am differentiating it now.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how about if the given is cos^x, then Yp=asin^2 x?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if the forcing term is cos^2 x you mean?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, for the same equation, I believe that a*sin^2 x would be the trial solution then...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes. it would. just checked it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0darn, i can't do it. show me a brief solution please. so i can learn how.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just for the Yp part. swear.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well, I did it already (a*cos^2(x))" +4a cos^2(x) = 4sin^2(x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.02a*cos^2(x)  2a*sin^2(x) +4a cos^2(x) = 4sin^2(x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm, I see your point ...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I guess you'll be needing something like yp = a*cos^2(x) + b*sin^2(x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0let me try it out and see...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0seems to work... I got a= 2 b=2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so 2cos^2 x 2sin^2 x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0answer here in the book is 1/2(1xsin2x) for the Yp part.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'll assume that's (1/2)*( 1  x*sin^2 x )

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you should check the original, I think you might have written the problem wrong...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you left out and x I think...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(1/2)*( sin^2 x  x*sin 2x ) ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't see how (1/2)(1xsin2x) can work... maybe I'm missing something.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0something's off in your original equation or that soln.
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