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Section 3.3 A Basic Example

Subsection 3.3.1 Solve \(\mathbf{x}'(t) = \begin{bmatrix} 2 \amp 3 \\ 4 \amp 1 \end{bmatrix} \mathbf{x}(t) + \begin{bmatrix} -e^{-t} \\ 2e^{-t} \end{bmatrix}\)

Let's label \(A = \begin{bmatrix} 2 \amp 3 \\ 4 \amp 1 \end{bmatrix}\text{.}\) First, we'll find the eigenvalues \(\lambda_1, \lambda_2\) of \(A\text{,}\) and the corresponding eigenvectors \(\mathbf{v}_1, \mathbf{v}_2\text{.}\)

You can find eigenvalues \(\lambda\) of \(A\) by solving the equation \(\det(A - \lambda I_2) = 0\) for \(\lambda\text{.}\)

\begin{align*} 0 = \det \begin{bmatrix} 2 - \lambda \amp 3 \\ 4 \amp 1 - \lambda \end{bmatrix} = (\lambda - 2) (\lambda - 1) - 12 = \lambda^2 -3\lambda - 10 = (\lambda + 2) (\lambda - 5). \end{align*}

Thus, the eigenvalues of \(A\) are \(\lambda_1 = -2\) and \(\lambda_2 = 5\text{.}\)

If \(\mathbf{v}\) is an eigenvector of \(A\) with eigenvalue \(\lambda\text{,}\) then \((A - \lambda I_2) \mathbf{v} = \mathbf{0}\text{.}\)

So, corresponding to \(\lambda_1 = -2\text{,}\) we find that

\begin{align*} (A - (-2) I_2) \mathbf{v} = \begin{bmatrix} 4 \amp 3 \\ 4 \amp 3 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}. \end{align*}

Hence, \(4v_1 + 3v_2 = 0\text{,}\) so the vector \(\mathbf{v}_1 = \bvec{-3}{4}\) is an eigenvector of \(A\) with eigenvalue \(\lambda_1 = -2\text{.}\)

By a similar process, we find that the vector \(\mathbf{v}_2 = \bvec{1}{-1}\) is an eigenvector of \(A\) with eigenvalue \(\lambda_2 = 5\text{.}\)

Here we've got eigenvalues \(\lambda_1 = -2\) and \(\lambda_2 = 5\text{.}\) Correspondingly, we have eigenvectors \(\mathbf{v}_1 = \bvec{-3}4\) and \(\mathbf{v}_2 = \bvec1{-1}\text{.}\)

So, we can write \(A = VDV^{-1}\) (ie. \(A\) is diagonalizable) where

\begin{gather*} V = [\mathbf{v}_1 \ \ \mathbf{v}_2] = \bmat{-3}{1}{4}{-1} \qquad \text{and} \qquad D = \bmat{-2}{0}{0}5 \end{gather*}

Following our steps, we write \(\mathbf x = V \mathbf y\text{,}\) so that we have

\begin{align*} V \mathbf y' \amp = AV \mathbf y + \bvec{-e^{-t}}{2e^{-t}}\\ \implies \mathbf y ' \amp = D \mathbf y + V^{-1} \bvec{-e^{-t}}{2e^{-t}} \end{align*}

Next, we should simplify \(V^{-1}\bvec{-e^{-t}}{2e^{-t}}\text{,}\) for which it turns out that

\begin{gather*} V^{-1} \bvec{-e^{-t}}{2e^{-t}} = \bvec{e^{-t}}{2e^{-t}}. \end{gather*}

First, we find \(V^{-1}\text{.}\)

\begin{align*} V^{-1} \amp = \frac1{\det V} \bmat{-1}{-1}{-4}{-3} = \bmat1143. \end{align*}

So,

\begin{gather*} V^{-1} \bvec{-e^{-t}}{2e^{-t}} = \bmat1143 \bvec{-1}{2} e^{-t} = \bvec{e^{-t}}{2e^{-t}}. \end{gather*}

Setting \(\mathbf y = \bvec{y_1}{y_2}\text{,}\) we now have to solve

\begin{gather*} \bvec{y_1'}{y_2'} = \bmat{-2}005 \bvec{y_1}{y_2} + \bvec{e^{-t}}{2e^{-t}}, \end{gather*}

which is equivalent to the linear system of ODEs

\begin{align*} y_1' \amp = -2y_1 + e^{-t}, \\ y_2' \amp = 5y_2 + 2e^{-t}. \end{align*}

Both of these can be solved using integrating factors.

For the ODE \(y_1' = -2y_1 + e^{-t}\text{,}\) first write

\begin{align*} y_1' + 2y_1 \amp = e^{-t}. \end{align*}

The expression \(e^{2t}\) is a good integrating factor since

\begin{gather*} e^{2t} y_1' + 2e^{2t} y_1 = [e^{2t} y_1]' \end{gather*}

So, multiplying the ODE by \(e^{2t}\text{,}\) we get

\begin{gather*} e^{2t} y_1' + 2e^{2t} y_1 = [e^{2t} y_1]' = e^t. \end{gather*}

Integrating gives us \(e^{2t} y_1 = e^t + c_1\text{,}\) for some constant \(c_1\text{.}\) Thus, the general solution for \(y_1\) is

\begin{align*} y_1(t) \amp = e^{-t} + c_1e^{-2t}. \end{align*}

By a similar process, we find that \(e^{-5t}\) is a good integrating factor for \(y_2' - 5y_2 = 2e^{-t}\text{,}\) so

\begin{gather*} [e^{-5t}y_2]' = 2e^{-6t} \quad\implies\quad e^{-5t}y_2 = \frac13 e^{-6t} + c_2. \end{gather*}

Hence, the general solution for \(y_2\) is

\begin{align*} y_2(t) \amp = \frac13 e^{-t} + c_2e^{5t}. \end{align*}

Using integrating factors gives us the general solutions of \(y_1\) and \(y_2\) separately:

\begin{align*} y_1(t) \amp = e^{-t} + c_1e^{-2t}, \\ y_2(t) \amp = \frac13 e^{-t} + c_2e^{5t}. \end{align*}

When we put these back into \(\mathbf y = \bvec{y_1}{y_2}\text{,}\) it might initially look a bit messy. But, separating the terms will give you a clearer picture of what the solution is!

\begin{align*} \mathbf y \amp = \bvec{e^{-t} + c_1e^{-2t}}{\frac13 e^{-t} + c_2e^{5t}},\\ \amp = c_1 \bvec{e^{-2t}}0 + c_2 \bvec0{e^{5t}} + \frac{e^{-t}}3 \bvec3{-1}. \end{align*}

Now we only need to calculate \(\mathbf x = V \mathbf y\text{.}\) Computing this is not too bad with the expanded form above.

\begin{align*} \mathbf x \amp = V \mathbf y = \bvec{-3}14{-1} \mathbf y \\ \amp = c_1 e^{-2t} \bvec{-3}14{-1} \bvec{1}0 + c_2 e^{5t} \bvec{-3}14{-1} \bvec0{1} + \frac{e^{-t}}3 \bvec{-3}14{-1} \bvec3{-1}\\ \amp = c_1 e^{-2t} \bvec{-3}4 + c_2 e^{5t} \bvec1{-1} + \frac{e^{-t}}3 \bvec{-8}{13}. \end{align*}

At last, we've found that the general solution \(\mathbf x(t)\) to \(\mathbf{x}'(t) = A \mathbf{x}(t) + \begin{bmatrix} -e^{-t} \\ 2e^{-t} \end{bmatrix}\) is

\begin{gather*} \boxed{\mathbf x(t) = c_1 e^{-2t} \bvec{-3}4 + c_2 e^{5t} \bvec1{-1} + \frac{e^{-t}}3 \bvec{-8}{13}}. \end{gather*}

Just looking at the form of the solution, the only condition we need to have is \(c_2 = 0\text{.}\)

Now, in the general solution, setting \(t = 0\) shows that

\begin{align*} \mathbf x(0) \amp = c_1 \bvec{-3}4 + c_2 \bvec11 + \bvec{-8/3}{13/3} = \bvec{-3c_1 + c_2 - 8/3}{4c_1 + c_2 + 13/3}. \end{align*}

Setting \(c_2 = 0\) and \(c_1 = k\) (just changing the name of \(c_1\)) shows that

\begin{align*} \mathbf x(0) \amp = k \bvec{-3}4 + \bvec{-8/3}{13/3}. \end{align*}

This completes the question, as it shows that \(\mathbf x(0)\) takes on values along a line.