Find the general solution?

Question

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.)

Answer 1

Hi Sun,

 

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

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

 

Did you get

x_h = c₁ (1,3) e^-t + c₂ (1,1) e^t ?

 

That's what the answer for this first part should be.
 
Now you need to deal with the (e^t, 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 x_h 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, e^t and t

in your case. That is, you assume x is of the form

 

x = A t + B e^t

 

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 e^t + 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) e^t, looks just like one of the terms in the inhomogeneous part, e^t. When that happens, there has to be another term in the particular solution, C te^t. Therefore, our starting point is

 

x = A t + B e^t + C t e^t + 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+(e^t, t)

 

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

Answer 2

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)=c₁e^-t(1,3)+c₂e^t(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)'e^t(1,1)+c₂(t)(e^t)'(1,1)={(2,3);(-1,-2)}[c₁(t)e^-t(1,3)+c₂(t)e^t(1,1)]+(e^t,t);

After multiplication in the right side, you will get:

c₁'(t)e^-t(1,3)+c₁(t)(e^-t)'(1,3)+c₂(t)'e^t(1,1)+c₂(t)(e^t)'(1,1)=-c₁(t)e^-t(1,3)+c₂(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:

c₁'(t)e^-t(1,3)-c₁(t)e^-t(1,3)+c₂(t)'e^t(1,1)+c₂(t)e^t(1,1)=

=-c₁(t)e^-t(1,3)+c₂(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:

c₁(t)'(1,3)e^-t+c₂(t)'(1,1)e^t=(e^t,t)

 

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

 

c₁'(t)=(te^t-e^2t)/2;

c₂'(t)=(3-te^-t)/2;

 

Integrating gives us:

c₁(t)=(t-1)e^t/2-e^2t/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:

 

x_inh=S₁e^-t(1,3)+S₂e^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=S₁e^-t(1,3)+S₂e^t(1,1)+(t-e^t/4+(3/2)te^t,2t-¾e^t+(3/2)te^t-1)