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# Find the general solution?

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.)

Take a look at my way of solving it. Is it clearer?

### 2 Answers by Expert Tutors

Kirill Z. | Physics, math tutor with great knowledge and teaching skillsPhysics, math tutor with great knowledge...
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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.

xh(t)=c1e-t(1,3)+c2et(1,1)

Now assume c1 and c2 are functions of t. Plug this back into the original equation. You will get:

c1'(t)e-t(1,3)+c1(t)(e-t)'(1,3)+c2(t)'et(1,1)+c2(t)(et)'(1,1)={(2,3);(-1,-2)}[c1(t)e-t(1,3)+c2(t)et(1,1)]+(et,t);
After multiplication in the right side, you will get:
c1'(t)e-t(1,3)+c1(t)(e-t)'(1,3)+c2(t)'et(1,1)+c2(t)(et)'(1,1)=-c1(t)e-t(1,3)+c2(t)et(1,1)+(et,t)
Take derivatives of e-t an et on the left side in the second and fourth terms, to obtain:
c1'(t)e-t(1,3)-c1(t)e-t(1,3)+c2(t)'et(1,1)+c2(t)et(1,1)=
=-c1(t)e-t(1,3)+c2(t)et(1,1)+(et,t)

You can see that two terms on the left side cancel two terms on the right side. (I underlined them)
We finally obtain:
c1(t)'(1,3)e-t+c2(t)'(1,1)et=(et,t)

This is the system of linear equations with respect to c1' and c2'. After solving it we obtain:

c1'(t)=(tet-e2t)/2;
c2'(t)=(3-te-t)/2;

Integrating gives us:
c1(t)=(t-1)et/2-e2t/4+S1;
c2(t)=[3t+(t+1)e-t]/2+S2;
Plug in back into the solution for the homogeneous system, you will obtain the following:

xinh=S1e-t(1,3)+S2et(1,1)+(2t-2-et)/4*(1,3)+(3tet+t+1)/2*(1,1)

Upon simple simplification, you will get:

xinh=S1e-t(1,3)+S2et(1,1)+(t-et/4+(3/2)tet,2t-¾et+(3/2)tet-1)

This method will give you a system of differential equations which is more complicated than the original system.
Not at all.
You're right, this system is not more complicated than the original system, because c1 and c2 cancel from the vector equations (which they don't when you use variation of parameters for scalar 2nd order equations). However: variation of parameters always involves taking integrals (with integration by parts), whereas undetermined coefficients does not. Anyway, I think Sun should be comfortable with both methods. :)
Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
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Hi Sun,

I'll try again. You wrote you know how to solve the homogeneous part,

xh'=(2, 3, -1, -2) xh .

Did you get

xh = c1 (1,3) e-t + c2 (1,1) et ?

That's what the answer for this first part should be.

Now you need to deal with the (et, t) part, and that's a little tricky in this case. You need to find a particular solution to your equation which is then added to the xh solution from above.

There are different ways to find a particular solution, but here is the simplest (called the method of undetermined coefficients, due to the Swiss mathematician Euler):

You assume that the particular solution contains terms that look just like the inhomogeneous terms, et and t
in your case. That is, you assume x is of the form

x = A t + B et

Here A and B are constants that we need to determine.

Now, whenever you have a t-term you could also have just a constant term, call it D=(D1,D2), so we need to modify our x equation to be

x = A t + B et + D.

Normally, you would now start to determine A, B, and D. In this problem, however, there is an additional difficulty: One of the two terms in the homogeneous solution, c2 (1,1) et, looks just like one of the terms in the inhomogeneous part, et. When that happens, there has to be another term in the particular solution, C tet. Therefore, our starting point is

x = A t + B et + C t et + D.

Since x is a vector with two components, A, B, C, and D must also be vectors. Let their components be given by

A=(A1, A2), B=(B1,B2),C=(C1,C2), and D=(D1,D2).

So we have 8 unknown constants that we now need to determine. We determine them by substituting x back into our original equation

x'=(2, 3, -1, -2)x+(et, t)

I need to start a new window to continue (?)

...continued.

Substitute x into your equation. Then separate the first and second component, get two equations. For each of those two equations, extract all terms that contain a t-term, all terms with an et term, all terms with a tet term, and all constant terms. You end up with 8 equations for the 8 unknowns. It takes some algebra which I show you if you would like me to to get the solutions:

A1 = 1, A2 = 2, B1  = -1/4, B2= -3/4, C1 = C2 = 3/2, D1 = 0, D2 = -1.

Therefore,

x = (t - 1/4 et + 3/2 tet, 2t - 3/4 et + 3/2 tet - 1)

Add this to xh=c1 (1,3) e-t + c2 (1,1) et to get the final answer,

x = (c1 e-t + c2 et+ t - 1/4 et + 3/2 tet, 3c1 e-t + c2 et+ 2t - 3/4 et + 3/2 tet - 1) .

I hope this helps!
But how do I do that using undetermined coefficients?
Sorry, but can you tell me how to plug the x into what equation? And how can I get all those values for A1, A2, etc.
You let x = A t + B et + C t et + D, so therefore x' = A + B et + C et + C t et . Let this be equal to

(2, 3, -1, -2) x + (et, t) = (2, 3, -1, -2) (A t + B et + C t et + D) + (et, t).

Let's look at the first component of this:

A1 + B1 et + C1 et + C1 t et
= 2(A1 t + B1 et + C1 t et + D1) +3 ( A2 t + B2 et + C2 t et + D2) + et
Collect all constants:

A1=2 D1 +3 D2

That's the first equation. Now collect all t terms:

0 = 2 A1 + 3 A2
That's the second equation. Continue until you have all 8 equations.
Do you remember seeing this method applied to second-order equations, like y''+y'+y = et ? It works the same way here.
Sorry, I switched rows and columns on the matrix. The first component is this:

A1 + B1 et + C1 et + C1 t et

= 2(A1 t + B1 et + C1 t et + D1) - ( A2 t + B2 et + C2 t et + D2) + et

Now collect the constants, t-terms etc.