I want to find a general solution of $$\ddot x-3\dot x+2x=e^t\sin(t)$$ At first, I looked at the differential equation $$\ddot x-3\dot x+2x=e^{(1+i)t}$$ whose imaginary part should be my desired solution. For the complex problem I found a special solution $$\varphi(t)=\left(-\frac{1}{2}+\frac{i}{2}\right)e^{(1+i)t}$$ Now, the characteristic polynomial $P(D)=D^2-3D+2$ of this differential equation has roots $\lambda_1=1$ and $\lambda_2=2$ so the fundamental system of solutions should be $$\left\{e^t,e^{2t}\right\}$$ Therefore, the general solution to the complex problem is $$\Phi(t)=\left(-\frac{1}{2}+\frac{i}{2}\right)e^{(1+i)t}+C_1e^t+C_2e^{2t}$$ where $C_1,C_2\in\mathbb C$ and I obtain the general solution of my original problem by $$\Psi(t)=\Im(\Phi(t))=\frac{e^t}{2}(\cos(t)-\sin(t))$$ and I don't think this can be right since there are no constants anywhere. Is there an error in my thoughts? And if so, where is it?
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$\begingroup$ You say $C_1$ and $C_2$ are complex no? $\endgroup$– PaulCommented May 31, 2018 at 9:45
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$\begingroup$ Oh, you're right! Is that it? $\endgroup$– BuhCommented May 31, 2018 at 9:47
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$\begingroup$ I'd say so. Try out an initial condition and find the (complex) constants $\endgroup$– PaulCommented May 31, 2018 at 9:48
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$\begingroup$ @Buh What do they mean $\ddot x,\, \dot x$ $\endgroup$– Darío A. GutiérrezCommented May 31, 2018 at 9:53
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$\begingroup$ @DaríoA.Gutiérrez it's notation for derivative of $x(t)$ in respect to time $t$. $\endgroup$– DigitalisCommented May 31, 2018 at 9:57
3 Answers
$$ e^{it} = \cos(t)+i\sin(t) $$
$$ (1+i)(\ddot x-3\dot x+2x)=e^t e^{it}\Rightarrow \ddot x-3\dot x+2x = \left(\frac{1-i}{2}\right)e^{t(i+1)} $$
The solution for the homogeneous equation or $\ddot x_h-3\dot x_h+2x_h = 0$ is
$$ x_h = C_1 e^t+C_2 e^{2t} $$
and a particular solution is
$$ x_p = \frac{e^t}{2}\left(\cos(t)-\sin(t)\right) $$
and then
$$ x = x_h + x_p = C_1 e^t+C_2 e^{2t}+\frac{e^t}{2}\left(\cos(t)-\sin(t)\right) $$
For a particular solution, try using $y=A e^t\sin t + B e^t \cos t $ and find the real numbers $A$ and $B$. Note that you get two equations in $A$ and $B$. Suppose you found them.
Hence, the general solution is
$$ x(t) = C_1 e^t + C_2 e^{2t} + A e^t\sin t + B e^t \cos t $$, where $C_1,C_2$ are arbitrary constants and $A$,B$ are known numbers.
$$\ddot x-3\dot x+2x=e^t\sin(t)$$ Another solution $$\ddot x-2\dot x- \dot x+2x=e^t\sin(t)$$ $$(\ddot x-2\dot x)-( \dot x-2x)=e^t\sin(t)$$ $$z'-z=e^t\sin(t)$$ $$z'e^{-t}-ze^{-t}=\sin(t)$$ Substitute $z=x'-2x$ $$(ze^{-t})'=\sin(t)$$ $$ze^{-t}=-\cos(t)+K$$ $$(x'-2x)e^{-t}=-\cos(t)+K$$ $$(xe^{-2t})'=-\cos(t)e^{-t}+Ke^{-t}$$ $$xe^{-2t}=-\int \cos(t)e^{-t}dt+Ke^{-t}+C$$ $$x(t)=-e^{2t}\int \cos(t)e^{-t}dt+Ke^{t}+Ce^{2t}$$ Therefore $$\boxed{x(t)=\frac 12e^{t}(\cos(t)-\sin(t))+Ke^{t}+Ce^{2t}}$$
