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