General solutions for non-homogeneous diff eq

Jun 2017
1
0
Romania
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 !
 

Country Boy

Math Team
Jan 2015
3,261
899
Alabama
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 \(\displaystyle y(s)= e^{2s}(A cos(s)+ B sin(s))+ (1/5)e^{2s}\).
 
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skipjack

Forum Staff
Dec 2006
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2,470
You slipped up; $P = 1$.