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February 12th, 2016, 10:22 PM   #1
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solution of differential equation

Find solution of
y" + 2y' + 5y =( 4cos2x)(e^-2)
If y(0)=y'(0)=0
Thanks for help.plz show the method also
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February 12th, 2016, 10:52 PM   #2
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Are you missing an $x$ in the exponential term?

The characteristic equation yields $m = -1 \pm 2\mathrm i$ so for the method of undetermined coefficients we may try $y_p = (A\cos 2x + B\sin 2x)\mathrm e^{-2x}$ because none of those terms exist in the complementary solution.
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February 13th, 2016, 12:14 AM   #3
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Yes, it is e^(-2x).

Please tell the solution... it's urgent...

Last edited by skipjack; February 15th, 2016 at 01:53 PM.
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February 14th, 2016, 08:33 AM   #4
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$\displaystyle y=e^{-3x}[\frac{2}{5}e^{x}(2\sin2x+\cos2x)+c]$

Last edited by skipjack; February 15th, 2016 at 01:51 PM.
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February 14th, 2016, 08:38 AM   #5
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Quote:
Originally Posted by rakmo View Post
Please tell the solution... it's urgent...
Perhaps you should put some effort in. Show some of your work.

Last edited by skipjack; February 15th, 2016 at 01:54 PM.
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February 15th, 2016, 10:02 AM   #6
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Thanks v8archie.
You helped a lot.
Because of you, I was able to figure it out and it helped me in exam.

Last edited by skipjack; February 15th, 2016 at 01:54 PM.
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February 15th, 2016, 12:40 PM   #7
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No problem.
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February 15th, 2016, 01:50 PM   #8
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According to W|A, y = e^(-2x)(2(9e^x - 8)sin(2x) - 4(e^x - 1)cos(2x))/17.
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