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Section 3.3 A Basic Example

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Subsection 3.3.1 Solve x′(t)=[2341]x(t)+[−e−t2e−t]

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Let's label A=[2341]. First, we'll find the eigenvalues λ1,λ2 of A, and the corresponding eigenvectors v1,v2.

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 \(\lambda_2 = 5\text{.}\)

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Here we've got eigenvalues λ1=−2 and λ2=5. Correspondingly, we have eigenvectors v1=[−34] and v2=[1−1].

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So, we can write A=VDV−1 (ie. A is diagonalizable) where

V=[v1  v2]=[−314−1]andD=[−2005]

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Following our steps, we write x=Vy, so that we have

Vy′=AVy+[−e−t2e−t]⟹y′=Dy+V−1[−e−t2e−t]

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Next, we should simplify V−1[−e−t2e−t], for which it turns out that

V−1[−e−t2e−t]=[e−t2e−t].

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*}

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Setting y=[y1y2], we now have to solve

[y1′y2′]=[−2005][y1y2]+[e−t2e−t],

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which is equivalent to the linear system of ODEs

y1′=−2y1+e−t,y2′=5y2+2e−t.

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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*}

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Using integrating factors gives us the general solutions of y1 and y2 separately:

y1(t)=e−t+c1e−2t,y2(t)=13e−t+c2e5t.

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When we put these back into y=[y1y2], it might initially look a bit messy. But, separating the terms will give you a clearer picture of what the solution is!

y=[e−t+c1e−2t13e−t+c2e5t],=c1[e−2t0]+c2[0e5t]+e−t3[3−1].

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Now we only need to calculate x=Vy. Computing this is not too bad with the expanded form above.

x=Vy=[−31]4−1y=c1e−2t[−31]4−1[10]+c2e5t[−31]4−1[01]+e−t3[−31]4−1[3−1]=c1e−2t[−34]+c2e5t[1−1]+e−t3[−813].

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At last, we've found that the general solution x(t) to x′(t)=Ax(t)+[−e−t2e−t] is

x(t)=c1e−2t[−34]+c2e5t[1−1]+e−t3[−813].

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.