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.)

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.)

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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} = c_{1} (1,3) e^{-t} + c_{2} (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**=(A_{1}, A_{2}), **B**=(B_{1},B_{2}),
**C**=(C_{1},C_{2}), and **D**=(D_{1},D_{2}). (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 A_{1} ... D_{2}. After a lot of algebra you will find

A_{1} = 1, A_{2} = 2, B_{1} = -1/4, B_{2}= -3/4, C_{1} = C_{2} = 3/2, D_{1} = 0, D_{2} = -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}.

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

Characteristic polynomial looks as follows:

(2-λ)(-2-λ)+3=0

λ^{2}-4+3=0

λ^{2}=1^{
}

λ_{1}=-1

λ_{2}=1

2) Find eigenvectors of A;

a) For λ_{1}=-1;

Solve {(3,3);(-1,-1)}(a_{1},a_{2})=0

You will get 3a_{1}-a_{2}=0 or (a_{1},a_{2})=(1,3)

First eigenvector is (1,3), corresponding to λ_{1}=-1.

b) For λ_{2}=1;

Solve {(1,3);(-1,-3)}(a_{1},a_{2})=0

You will get a_{1}-a_{2}=0 or (a_{1},a_{2})=(1,1)

Second eigenvector is (1,1), corresponding to λ_{2}=1.

Solve {(1,3);(-1,-3)}(a

You will get a

Second eigenvector is (1,1), corresponding to λ

The solution of a homogenous system looks like:

In order to solve inhomogenous system, we may now assume that c_{1} and c_{2} depend on t and plug this into the inhomogenous equation.

We will get:

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

Some terms cancel and we obtain:

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

Solving for c_{1}(t) and c_{2}(t) yields:

c_{1}(t)=(t-1)e^{t}/2-t/2+C_{1};

c_{2}(t)=t(1+e^{-t}/2)+te^{-2t}/4+e^{-2t}/8+C_{2};

Thus the general solution is:

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