Find the general solution of x'=(2, 3, -1, -2)x+(e^t, t). (this is 2x2 matrix with 2 and 3 on left, -1 and -2 on right. And I know how to find the general solution for the 2x2 matrix but not the (e^t, t) part. Can someone please help me step by step? the
e^t is on top, t is on bottom.)

Use the method of variation of parameters. Once the solution to a homogeneous system has been found, it has two arbitrary constants. Assume that they are functions of t, plug in the solution into inhomogeneous equation, then you will obtain a system of
linear equations for the first-order derivatives of those variable "Constants". Solve it and integrate to get variables themselves. Plug in into original solution of homogeneous equation and the solution of inhomogeneous equation is obtained.

Let me illustrate:

The solution to a homogeneous system is given by Andre.

x

_{h}(t)=c_{1}e^{-t}(1,3)+c_{2}e^{t}(1,1)Now assume c

_{1}and c_{2}are functions of t. Plug this back into the original equation. You will get:c

_{1}'(t)e^{-t}(1,3)+c_{1}(t)(e^{-t})'(1,3)+c_{2}(t)'e^{t}(1,1)+c_{2}(t)(e^{t})'(1,1)={(2,3);(-1,-2)}[c_{1}(t)e^{-t}(1,3)+c_{2}(t)e^{t}(1,1)]+(e^{t},t);After multiplication in the right side, you will get:

c

_{1}'(t)e^{-t}(1,3)+c_{1}(t)(e^{-t})'(1,3)+c_{2}(t)'e^{t}(1,1)+c_{2}(t)(e^{t})'(1,1)=-c_{1}(t)e^{-t}(1,3)+c_{2}(t)e^{t}(1,1)+(e^{t},t)Take derivatives of e

^{-t}an e^{t}on the left side in the second and fourth terms, to obtain:c

_{1}'(t)e^{-t}(1,3)-c_{1}(t)e^{-t}(1,3)+c_{2}(t)'e^{t}(1,1)+c_{2}(t)e^{t}(1,1)==-c

_{1}(t)e^{-t}(1,3)+c_{2}(t)e^{t}(1,1)+(e^{t},t)You can see that two terms on the left side cancel two terms on the right side. (I underlined them)

We finally obtain:

c

_{1}(t)'(1,3)e^{-t}+c_{2}(t)'(1,1)e^{t}=(e^{t},t)This is the system of linear equations with respect to c

_{1}' and c_{2}'. After solving it we obtain:c

_{1}'(t)=(te^{t}-e^{2t})/2;c

_{2}'(t)=(3-te^{-t})/2;Integrating gives us:

c

_{1}(t)=(t-1)e^{t}/2-e^{2t}/4+S_{1};c

_{2}(t)=[3t+(t+1)e^{-t}]/2+S_{2};^{ }Plug in back into the solution for the homogeneous system, you will obtain the following:

x

_{inh}=S_{1}e^{-t}(1,3)+S_{2}e^{t}(1,1)+(2t-2-e^{t})/4*(1,3)+(3te^{t}+t+1)/2*(1,1)Upon simple simplification, you will get:

x

_{inh}=S_{1}e^{-t}(1,3)+S_{2}e^{t}(1,1)+(t-e^{t}/4+(3/2)te^{t},2t-¾e^{t}+(3/2)te^{t}-1)
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