Nonhomogeneous System General Solution Special Proviso

Question

I've gotten as far as to guess the particular solution as in the form

xp= ae^2t+bte^2t+ct^2*e^2t

so that the solution does not coincide with the first one

However, I'm stuck with matching the coefficients. Thanks.

Answer 1

I would solve this particular system by solving the second equation first, since it is purely in $x_2(t)$.

$$x_2'-2x_2=e^{2t}$$

It has homogeneous part $x_{2h}=C_2e^{2t}$ We look for a particular part for it of the form $x_{2p}=Ate^{2t}$. Then we have $x_{2p}'=Ae^{2t}+2Ate^{2t}$. $$x_{2p}'-2x_{2p}=2e^{2t}\implies Ae^{2t}=2e^{2t}\implies A=2$$ So we have $x_2=(2t+C_2)e^{2t}$.

Now we proceed with $x_1$. $$x_1'=2x_1+x_2+e^{2t}\implies x_1'-2x_1=(2t+C_2+1)e^{2t}$$ This has homogeneous part $x_{1h}=C_1e^{2t}$

We look for a particular part of the form $x_{1p}=(At^2+Bt)e^{2t}$. We have $x_{1p}'=(2At+B)e^{2t}+2(At^2+Bt)e^{2t}$. $$x_{1p}'-2x_{1p}=(2t+C_2+1)e^{2t}\implies(2At+B)e^{2t}=(2t+C_2+1)e^{2t}\implies\{A=1,B=C_2+1\}$$ So we have $x_1=(t^2+(C_2+1)t+C_1)e^{2t}$

In vector form:

$$\mathbf{x}= \begin{bmatrix} t^2+(C_2+1)t+C_1\\ 2t+C_2 \end{bmatrix}e^{2t}$$