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June 4th, 2017, 11:13 AM   #1
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Question 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 !
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June 4th, 2017, 01:11 PM   #2
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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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Last edited by skipjack; June 5th, 2017 at 12:28 AM.
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June 5th, 2017, 12:25 AM   #3
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You slipped up; $P = 1$.
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