
Differential Equations Ordinary and Partial Differential Equations Math Forum 
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February 12th, 2016, 10:22 PM  #1 
Member Joined: Nov 2012 Posts: 61 Thanks: 0  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 
February 12th, 2016, 10:52 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,696 Thanks: 2681 Math Focus: Mainly analysis and algebra 
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. 
February 13th, 2016, 12:14 AM  #3 
Member Joined: Nov 2012 Posts: 61 Thanks: 0 
Yes, it is e^(2x). Please tell the solution... it's urgent... Last edited by skipjack; February 15th, 2016 at 01:53 PM. 
February 14th, 2016, 08:33 AM  #4 
Senior Member Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math 
$\displaystyle y=e^{3x}[\frac{2}{5}e^{x}(2\sin2x+\cos2x)+c]$
Last edited by skipjack; February 15th, 2016 at 01:51 PM. 
February 14th, 2016, 08:38 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,696 Thanks: 2681 Math Focus: Mainly analysis and algebra  Perhaps you should put some effort in. Show some of your work.
Last edited by skipjack; February 15th, 2016 at 01:54 PM. 
February 15th, 2016, 10:02 AM  #6 
Member Joined: Nov 2012 Posts: 61 Thanks: 0 
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. 
February 15th, 2016, 12:40 PM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,696 Thanks: 2681 Math Focus: Mainly analysis and algebra 
No problem.

February 15th, 2016, 01:50 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 21,105 Thanks: 2324 
According to WA, y = e^(2x)(2(9e^x  8)sin(2x)  4(e^x  1)cos(2x))/17.


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