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 June 4th, 2017, 10:13 AM #1 Newbie   Joined: Jun 2017 From: Romania Posts: 1 Thanks: 0 Math Focus: All of it General solutions for non-homogeneous diff eq Hey guys ! I have some troubles with these types of differential equations. I have to find the general solution of this eq : y''-4y'+5y=e^(2s) I have found the general solution of the homogeneous part of this eq. Yh= e^(2s) * ( c1* cos s - c2 * sin s ) I hope it's correct. Well, my problem comes at the particular solution. I don't understand how to find it. Can anyone help me? Thank you !
 June 4th, 2017, 12:11 PM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,198 Thanks: 872 Yes, that is the correct solution to the "associated homogeneous equation". I presume you know that the general solution to the entire equation is that general solution to the associated homogeneous equation plus a single solution to the entire equation. One method for finding that one solution, called "undetermined coefficients" is to look for a solution of the form $\displaystyle Pe^{2s}$ (we need to "determine" the coefficient P). If $\displaystyle y= Pe^{2s}$ then $\displaystyle y'= 2Pe^{2s}$ and $\displaystyle y''= 4Pe^{2s}$ so that $\displaystyle y''- 4y'+ 5y= 4Pe^{2s}- 4Pe^{2s}+ 5Pe^{2s}= 5Pe^{2s}= e^{2s}$ so that P= 1/5. The general solution to the differential equation is $\displaystyle y(s)= e^{2s}(A cos(s)+ B sin(s))+ (1/5)e^{2s}$. Thanks from romsek Last edited by skipjack; June 4th, 2017 at 11:28 PM.
 June 4th, 2017, 11:25 PM #3 Global Moderator   Joined: Dec 2006 Posts: 19,062 Thanks: 1619 You slipped up; $P = 1$.

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