Answer to Question #239627 in Differential Equations for Charles

Question

Use the method of variation of parameters to solve the system

𝑋′ =[

3 −1 −1

−2 3 2

4 −1 −2

]𝑋+[

1

𝑒𝑡

𝑒𝑡]

Answer 1

The given question is not at all clear. So, I am taking a example and solving that.

Question: Find a particular solution of:

$\mathbf{y}^{\prime}=\left[\begin{array}{ccc} 2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0 \end{array}\right] \mathbf{y}+\left[\begin{array}{c} e^{t} \\ 0 \\ e^{-t} \end{array}\right]$

Solution:

The characteristic polynomial of the coefficient matrix is

$\left|\begin{array}{lll} 2-\lambda & -1 & -1 \\ 1 & -\lambda & -1 \\ 1 & -1 & -\lambda \end{array}\right|=-\lambda(\lambda-1)^{2}$

$\mathbf{y}_{1}=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right],\quad \mathbf{y}_{2}=\left[\begin{array}{l} e^{t} \\ e^{t} \\ 0 \end{array}\right],\quad \text { and } \quad \mathbf{y}_{3}=\left[\begin{array}{l} e^{t} \\ 0 \\ e^{t} \end{array}\right]$

are linearly independent solutions.

$Y=\left[\begin{array}{lll} 1 & e^{t} & e^{t} \\ 1 & e^{t} & 0 \\ 1 & 0 & e^{t} \end{array}\right]$ is a fundamental matrix. We seek a particular solution

$\mathbf{y}_{p}=Y \mathbf{u}$ where $Y \mathbf{u}^{\prime}=\mathbf{f}$ ; that is:

$\left[\begin{array}{lll} 1 & e^{t} & e^{t} \\ 1 & e^{t} & 0 \\ e & 0 & e^{t} \end{array}\right]\left[\begin{array}{l} u_{1}^{\prime} \\ u_{2}^{\prime} \\ u_{3}^{\prime} \end{array}\right]=\left[\begin{array}{l} e^{t} \\ 0 \\ e^{-t} \end{array}\right]$

The determinant of $Y$  is the Wronskian:

$\left|\begin{array}{lll} 1 & e^{t} & e^{t} \\ 1 & e^{t} & 0 \\ 1 & 0 & e^{t} \end{array}\right|=-e^{2 t}$

Thus, by Cramer's rule,

$\begin{aligned} u_{1}^{\prime} &=-\frac{1}{e^{2 t}}\left|\begin{array}{lll} e^{t} & e^{t} & e^{t} \\ 0 & e^{t} & 0 \\ e^{-t} & 0 & e^{t} \end{array}\right| &=-\frac{e^{3 t}-e^{t}}{e^{2 t}}=e^{-t}-e^{t} \\ u_{2}^{\prime} &=-\frac{1}{e^{2 t}}\left|\begin{array}{ccc} 1 & e^{t} & e^{t} \\ 1 & 0 & 0 \\ 1 & e^{-t} & e^{t} \end{array}\right| &=-\frac{1-e^{2 t}}{e^{2 t}}=1-e^{-2 t} \end{aligned}$

$u_{3}^{\prime}=-\frac{1}{e^{2 t}}\left|\begin{array}{ccc} 1 & e^{t} & e^{t} \\ 1 & e^{t} & 0 \\ 1 & 0 & e^{-t} \end{array}\right|=\frac{e^{2 t}}{e^{2 t}} \quad=1$

Therefore

$\mathbf{u}^{\prime}=\left[\begin{array}{l} e^{-t}-e^{t} \\ 1-e^{-2 t} \\ 1 \end{array}\right]$

Integrating and taking the constants of integration to be zero yields

$\mathbf{u}=\left[\begin{array}{l} -e^{t}-e^{-t} \\ \frac{e^{-2 t}}{2}+t \\ t \end{array}\right] \\ \mathbf{y}_{p}=Y \mathbf{u}=\left[\begin{array}{lll} 1 & e^{t} & e^{t} \\ 1 & e^{t} & 0 \\ 1 & 0 & e^{t} \end{array}\right]\left[\begin{array}{l} -e^{t}-e^{-t} \\ \frac{e^{-2 t}}{2}+t \\ t \end{array}\right]=\left[\begin{array}{c} e^{t}(2 t-1)-\frac{e^{-t}}{2} \\ e^{t}(t-1)-\frac{e^{-t}}{2} \\ e^{t}(t-1)-e^{-t} \end{array}\right]$ is a particular solution.