Given $x''+3x'+2x=4.$ ($''=2nd $ derivative, $'=1st$ derivative)
Determine the characteristic equation of this differential equation.
I'm having a hard time doing this because of that $4$. Any help would be appreciated. Thanks.
$\endgroup$ 101 Answer
$\begingroup$The $4$ is the inhomogeneous part of the equation. To evaluate the characteristic equation you have to consider only the homogeneous part:
$x^{\prime\prime} + 3 x^\prime + 2 x = 0$.
The characteristic equation, expressed in terms of a variable $\alpha$, is
$\alpha^2 + 3 \alpha + 2 = 0$.
The solutions are $\alpha = -2$ and $\alpha = -1$.
From this, you can obtain the solution of the homogeneous equation:
$x_h = A e^{-t} + B e^{-2t}$,
where $A$ and $B$ are arbitrary constants that you may probably have to fix using initial conditions.
The particular solution of the inhomogeneous equation is pretty simple because the inhomogeneous term is just a constant:
$x_p = 2$.
Therefore, the general solution of the equation is
$x = A e^{-t} + B e^{-2t} + 2$.
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