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.
$$Y_h= e^{2s} ( C_1 \cos s - C_2 \sin s ) $$
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 !
$\endgroup$ 22 Answers
$\begingroup$This is more a comment than an answer.
I think that life would have been easier from the beginning defining $y=e^{2s} z$ making the equation to become $$z''+z=1$$ Now, defining $z=w+1$ makes $$w''+w=0$$ giving, as usual $$w=c_1 \cos (s)+c_2 \sin (s)$$ from which $$z=1+c_1 \cos (s)+c_2 \sin (s)$$ $$y=e^{2s}\left(1+c_1 \cos (s)+c_2 \sin (s) \right)$$
$\endgroup$ $\begingroup$Hint:
To find the particular solution $Y_p$, let $y_p = Ce^{2s}$ and substitute in the differential equation to find the value of $C$.
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