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Question

Find the general solution of x'=(2, 3, -1, -2)x+(e^t, t). (2 and 3 on the left, -1 and -2 on the right. e^t on the top, t on the bottom.)

 

 

Answer 1

1) Find eigenvalues of matrix A={(2,3);(-1,-2)}

 

Characteristic polynomial looks as follows:

(2-λ)(-2-λ)+3=0 

λ²-4+3=0

λ²=1

λ₁=-1

λ₂=1

 

2) Find eigenvectors of A;

 

a) For λ₁=-1; 

Solve {(3,3);(-1,-1)}(a₁,a₂)=0

You will get 3a₁-a₂=0 or (a₁,a₂)=(1,3)

First eigenvector is (1,3), corresponding to λ₁=-1.

 

b) For λ₂=1;
Solve {(1,3);(-1,-3)}(a₁,a₂)=0
You will get a₁-a₂=0 or (a₁,a₂)=(1,1)
Second eigenvector is (1,1), corresponding to λ₂=1.

 

The solution of a homogenous system looks like:

 

x(t)=c₁(1,3)e^-t+c₂(1,1)e^t.

 

In order to solve inhomogenous system, we may now assume that c₁  and c₂ depend on t and plug this into the inhomogenous equation.

 

We will get:

 

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

 

Some terms cancel and we obtain:

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

 

Solving for c₁(t) and c₂(t) yields:

 

c₁(t)=(t-1)e^t/2-t/2+C₁;

c₂(t)=t(1+e^-t/2)+te^-2t/4+e^-2t/8+C₂;

 

Thus the general solution is:

 

x(t)=C₁(1,3)e^-t+C₂(1,1)e^t+[(t-1)/2-te^-t/2](1,3)+[te^t+t/2+¼*(t+½)e^-t](1,1)

 

Answer 2

This is an inhomogeneous system, with the inhomogeneity being (e^t,t).

 

The general solution will be the sum of the general solution to the homogeneous system,

 

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

 

and any particular solution to the inhomogeneous system, x_p :

 

x = x_h + x_p

 

You solve the homogeneous system as usual by finding the eigenvalues and eigenvectors first. The eigenvalues of the matrix (2, 3, -1, -2) are -1 and 1, the corresponding eigenvectors are (1,3) and (1,1).

 

Therefore,

 

x_h = c₁ (1,3) e^-t + c₂ (1,1) e^t .

 

To find a particular solution, you should use the method of undetermined coefficients. Let me know if you are unfamiliar with this method. You assume that x_p is of a form similar to your inhomogeneous term (e^t, t), namely

 

x_p = A t + B e^t + C te^t + D

 

Here A, B, C, and D are the undetermined coefficients that we determine by substituting this solution into the original equation. They are vectors with components A=(A₁, A₂), B=(B₁,B₂), C=(C₁,C₂), and D=(D₁,D₂). (Don't confuse the big C's with the little c's from the homogeneous solution.) Substitute all of this into your equation. Separate out the upper and lower equations for t, e^t, te^t, and 1, and get 8 equations for the 8 unknowns A₁ ... D₂. After a lot of algebra you will find

 

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

 

Therefore,

 

x_p = (t - 1/4 e^t + 3/2 te^t, 2t - 3/4 e^t + 3/2 te^t - 1)

 

Add this to x_h to get the final answer, x = x_h + x_p.

 

 

 

 

Answer 3

Can you clarify?