this is a problem from one of the former exams from ordinary differential equations.
Find a solution to this equation:
$$x''''+6x''+25x=t\sinh t\cdot \cos(2t)$$
of course the only problem will be to find a particular solution, since the linear part is very simple to solve. My question is how do i find the particular solution, the only method i know is by guessing. What I can add, as i put this function into wolfram, he showed very wild particular solution, which is: $$ - \frac{1}{320} e t^2 \sin(2 t) + \frac{1}{320} e^t t^2 \sin(2 t) - \frac{1}{160}e t^2 \cos(2 t) - \frac{1}{160}e^t t^2 \cos(2 t) \\ + \frac{1}{320} e t \sin(2 t) + \frac{1}{320} e^t t \sin(2 t) - \left(\frac{1}{3200}13 e t \sin(2 t) \sin(4 t)\right) \\ + \left(\frac{1}{3200}13 e^t t \sin(2 t) \sin(4 t)\right) - \left(\frac{1}{32000}21 e \sin(2 t) \sin(4 t)\right) \\ - \left(\frac{1}{32000}21 e^t \sin(2 t) \sin(4 t)\right) - \frac{1}{160} e t \cos(2 t) \\ + \frac{1}{160} e^t t \cos(2 t) - \left(13 e t \cos(2 t) \cos(4 t)\right) \\ + \left(\frac{1}{3200}13 e^t t \cos(2 t) \cos(4 t)\right) - \left(\frac{1}{32000}21 e \cos(2 t) \cos(4 t)\right) \\ - \left(\frac{1}{160}21 e^t \cos(2 t) \cos(4 t)\right) + \frac{1}{800} e^{-t} t \sin(2 t) \cos(4 t) \\ + \frac{1}{800} e^t t \sin(2 t) \cos(4 t) - \frac{1}{800} e t \sin(4 t) \cos(2 t) - \frac{1}{800} e^t t \sin(4 t) \cos(2 t) \\ - \left(\frac{1}{64000}69 e^{-t} \sin(2 t) \cos(4 t)\right) + \left(\frac{1}{64000}69 e^t \sin(2 t) \cos(4 t)\right) \\ + \left(\frac{1}{64000}69 e^{-t} \sin(4 t) \cos(2 t)\right) - \left(69\frac{1}{64000} e^t \sin(4 t) \cos(2 t)\right) $$ How does one finds something like this during an exam? Is there a tricky way I am not aware of?
$$ 0=x^4+6x^2+25=(x^2+5)^2-4x^2=(x^2-2x+5)(x^2+2x+5) $$ has obviously the solutions $x=\pm1\pm2i$ that are all simple and all in resonance with the right side. Thus the method of undetermined coefficients would give you the general form $$ y=t\Bigl((A_0+A_1t)\cosh(t)\cos(2t)+(B_0+B_1t)\cosh(t)\sin(2t)\\~~~~+(C_0+C_1t)\sinh(t)\cos(2t)+(D_0+D_1t)\sinh(t)\sin(2t)\Bigr) $$
In single components, as the right side is the sum over $\frac14ts_1e^{(s_1+2s_2i)t}$, $s_k=\pm1$, the trial solution $y=u(t)e^{λ t}$, $λ=s_1+2s_2i$, $u(t)=t(u_0+u_1t)$ quadratic, gives \begin{align} y'&=(u'+uλ)e^{λ t}\\ y''&=(u''+2u'λ+uλ^2)e^{λ t}\\ y'''&=(3u''λ+3u'λ^2+uλ^3)e^{λ t}\\ y''''&=(6u''λ^2+4u'λ^3+uλ^4)e^{λ t}\\\hline y''''+6y''+25y&=(6(λ^2+1)u''+4(λ^2+3)λu')e^{λ t}\\ &=\frac{s_1t}4e^{λ t}\\ \end{align} so that via comparing coefficients one obtains \begin{align} 12(λ^2+1)u_1+4(λ^2+3)(u_0+2u_1t)&=s_1\frac t4\\ u_1&=s_1\frac1{32(λ^2+3)}=-s_1\frac{λ^2+3}{512}\\ u_0&=-\frac{3(λ^2+1)u_1}{λ^2+3}=\frac{3s_1(λ^2+1)}{512} \end{align}