# Non homogeneous linear system

simmonj7

## Homework Statement

Verify that the given vector is the general solution of the corresponding homogeneous system and then solve the nonhomogenous system. Assume that t>0.

tx' =
|2 -1|x + |1- t^2|
|3 -2| |2t |

General solution:
x =
c1| 1|t + c2| 1|t^-1
| 1| |3|

This won't show up correctly but the first eigen vector is (1,1) and the second is (1,3)

## The Attempt at a Solution

So I have been solving non homogeneous linear systems all night with no problems however when I got to this problem I got stumped because there is a t in front of x'. I solved for the eigen values and eigen vectors and the eigen vectors match what they have but I have the general solution by solving it without acknowledging the t in front of x'. I thought that maybe I could just divide my general solution I got by t and then I would have the same answer, however that doesn't work because the solution provided by the problem doesn't have any e^t's in it. So I am not quite sure on what is going on exactly.

Homework Helper
hi simmonj7! (use the CODE tag for matrices … it's not what it's there for, but it does work! )

for the homogeous solution, i suspect you're meant to assume a solution of the form tn

for the particular solution, if it wasn't a matrix, you'd try a polynomial P(t) …

so try a vector of two polynomials, (P(t),Q(t)) Homework Helper
The problem is
$$t\begin{bmatrix}x \\ y\end{bmatrix}'= \begin{bmatrix}2 & -1 \\ 3 & -2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}+ \begin{bmatrix} 1- t^2 \\ 2t\end{bmatrix}$$

and you want to show that
$$c_1\begin{bmatrix}1 \\ 1\end{bmatrix}t+ c_2\begin{bmatrix}1 \\ 3\end{bmatrix}t^{-1}$$
is the general solution.

Both you and tiny-tim talk about "solving" the system but you are not asked to solve it! You are only asked to verify that the given solution works and that is a far easier problem.
(Which would be easier, to solve $x^5- 3x^4+ 4x^2- 3x+ 1= 0$ or to show that x= 1 is a solution?)

To show that the given function is the general solution, you need to use the fact that the general solution of a linear equation is a general linear combination of two independent solutions.

So you need to show three things:
1) That
$$\begin{bmatrix}1 \\ 1\end{bmatrix}t= \begin{bmatrix}t \\ t\end{bmatrix}$$
satisfies the equation by putting it and its derivative into the equation.

2) That
$$\begin{bmatrix}1 \\ 3\end{bmatrix}t^{-1}= \begin{bmatrix}t^{-1} \\ 3t^{-1}\end{bmatrix}$$
satisfies the equation by putting it and its derivative into the equation.

3) That those two functions are independent (that one is not a multiple of the other).