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.)

Tutors, please sign in to answer this question.

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:

Can you clarify?

Jing X.

IVY-graduate specializing in SAT, ACT, Math and College App

Brooklyn, NY

4.9
(76 ratings)

Jennifer C.

Experienced Educator Specializing in Test Prep and Enrichment

Brooklyn, NY

5.0
(70 ratings)

Amaan M.

Math/Economics Teacher

New York, NY

5.0
(68 ratings)