Question 3.12.2: Solve x' - 4x + y" = t² x' + x + y' = 0. (9)...

First we write the system in differential operator notation:

$\begin{aligned} (D-4) x+D^{2} y & =t^{2} \\ (D+1) x+D y & =0. \end{aligned}$    (10)

Then, by eliminating $x$, we obtain

$\left[(D+1) D^{2}-(D-4) D\right] y=(D+1) t^{2}-(D-4) 0$

or                        $\left(D^{3}+4 D\right) y=t^{2}+2 t .$

Since the roots of the auxiliary equation $m\left(m^{2}+4\right)=0$ are $m_{1}=0, m_{2}=2 i$, and $m_{3}=-2 i$, the complementary function is

$y_{c}=c_{1}+c_{2} \cos 2 t+c_{3} \sin 2 t.$

To determine the particular solution $y_{p}$ we use undetermined coefficients by assuming $y_{p}=A t^{3}+B t^{2}+C t$. Therefore

The last equality implies $12 A=1,8 B=2,6 A+4 C=0$, and hence $A=\frac{1}{12}, B=\frac{1}{4}, C=-\frac{1}{8}$. Thus

$y=y_{c}+y_{p}=c_{1}+c_{2} \cos 2 t+c_{3} \sin 2 t+\frac{1}{12} t^{3}+\frac{1}{4} t^{2}-\frac{1}{8} t.$   (11)

Eliminating $y$ from the system (9) leads to

$[(D-4)-D(D+1)] x=t^{2} \quad \text { or } \quad\left(D^{2}+4\right) x=-t^{2}.$

It should be obvious that

$x_{c}=c_{4} \cos 2 t+c_{5} \sin 2 t$

and that undetermined coefficients can be applied to obtain a particular solution of the form $x_{p}=A t^{2}+B t+C$. In this case the usual differentiations and algebra yield $x_{p}=-\frac{1}{4} t^{2}+\frac{1}{8}$, and so

$x=x_{c}+x_{p}=c_{4} \cos 2 t+c_{5} \sin 2 t-\frac{1}{4} t^{2}+\frac{1}{8}$.    (12)

Now $c_{4}$ and $c_{5}$ can be expressed in terms of $c_{2}$ and $c_{3}$ by substituting (11) and (12) into either equation of (9). By using the second equation, we find, after combining terms,

$\left(c_{5}-2 c_{4}-2 c_{2}\right) \sin 2 t+\left(2 c_{5}+c_{4}+2 c_{3}\right) \cos 2 t=0$

so that $c_{5}-2 c_{4}-2 c_{2}=0$ and $2 c_{5}+c_{4}+2 c_{3}=0$. Solving for $c_{4}$ and $c_{5}$ in terms of $c_{2}$ and $c_{3}$ gives $c_{4}=-\frac{1}{5}\left(4 c_{2}+2 c_{3}\right)$ and $c_{5}=\frac{1}{5}\left(2 c_{2}-4 c_{3}\right)$. Finally, a solution of $(9)$ is found to be