# Lecture 7: Systems of Differential Equations

### Instructions

• This section covers the concepts listed below. You can click the button to go directly to that topic.
• After each video, there are notes for the material in the video.
• At the end of the page, there are exercises covering the material in the section.
• When you have finished the material below, you can go to the next section or return to the main page.

### Concepts

Systems of Differential Equations

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#### Video Notes

The scalar–valued initial value problem:
\begin{align*}
x' = mx, \hspace{1in} x(0) = c
\end{align*}
has solution:
\begin{align*}
x(t) = e^{mt}c.
\end{align*}
As you have seen (or will see), systems of linear differential equations can be written in the form:
\begin{align*}
\begin{pmatrix}x_1'(t)\\\vdots\\x_n'(t)\end{pmatrix}
= M\begin{pmatrix}x_1(t)\\\vdots\\x_n(t)\end{pmatrix}.
\end{align*}
Or, more compactly as
\begin{align*}
\boldsymbol{x}' = M \boldsymbol{x}.
\end{align*}
If we reason in analogy with the scalar case, the solution to this equation is
\begin{align*}
\boldsymbol{x}(t) = e^{Mt}\boldsymbol{x}(0),
\end{align*}
Of course, we don't know what $$e^{Mt}$$ is k or how to make sense of it. Doing that is one of the goals of this lecture.

First, in the scalar case, we know that the general solution to the equation $$x'(t) = mx(t)$$ is $$x(t) = ce^{mt}$$. We are comfortable with the
function $$e^{mt}$$ so we don't think too much about what $$e^{mt}$$ means. There are actually several ways to define the exponential function. One way is to define it as the unique solution to $$x' = mx$$ with $$x(0)=1$$. The other way is to define it using power series:
\begin{align*}
e^{mt}
=\sum_{k=0}^{\infty}\frac{(mt)^k}{k!}.
\end{align*}
Observe that:
\begin{align*}
\frac{d}{dt}e^{mt}
&=\sum_{k=0}^{\infty}k\frac{m^k t^{k-1}}{k!}
\\[2mm]&=\sum_{k=1}^{\infty}\frac{m^{k}t^{k-1}}{(k-1)!}
\\[2mm]&=\sum_{k=0}^{\infty}\frac{m^{k+1}t^{k}}{k!}
\\[2mm]&=m e^{mt}.
\end{align*}
That is, we have verified that $$x(t)=e^{mt}$$ satisfies $$x' = mx$$.

Additionally, if $$x' = mx$$, then a straight–forward computation shows that $$x^{(k)}(0) = \frac{m^k}{k!}$$ and so
\begin{align*}
x(t) = \sum_{k=0}^{\infty}\frac{m^k}{k!}t^k.
\end{align*}
That is, both definitions are equivalent.

​To see the full video page and find related videos, click the following link.

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#### Video Notes

In the case of matrices, we can also define $$e^{Mt}$$ using series:
\begin{align*}
e^{Mt}
=\sum_{k=0}^{\infty} \frac{M^k}{k!}t^k.
\end{align*}
In general, this is a difficult sum to compute or to analyze. However, if $$M$$ has $$n$$ linearly independent eigenvectors $$\boldsymbol{v}_1,\ldots,\boldsymbol{v}_n$$ with corresponding eigenvalues $$\lambda_1,\ldots,\lambda_n$$ then we say in the previous
video, we can factor $$M$$ as:
\begin{align*}

=\begin{pmatrix}\boldsymbol{v}_1&\cdots&\boldsymbol{v}_n\end{pmatrix}
D\begin{pmatrix}\boldsymbol{v}_1&\cdots&\boldsymbol{v}_n\end{pmatrix}^{-1},
\end{align*}
where $$D$$ is the diagonal matrix
\begin{align*}
\begin{pmatrix}
\lambda_1 & 0 & \cdots & 0\\
0 & \lambda_2 & \cdots & 0\\
0 & 0 & \ddots & 0\\
0 & 0 & \cdots & \lambda_n
\end{pmatrix}.
\end{align*}
We also saw in the previous lecture that:
\begin{align*}
\begin{pmatrix}
\lambda_1 & 0 & \cdots & 0\\
0 & \lambda_2 & \cdots & 0\\
0 & 0 & \ddots & 0\\
0 & 0 & \cdots & \lambda_n
\end{pmatrix}^k
=\begin{pmatrix}
\lambda_1^k & 0 & \cdots & 0\\
0 & \lambda_2^k & \cdots & 0\\
0 & 0 & \ddots & 0\\
0 & 0 & \cdots & \lambda_n^k
\end{pmatrix}.
\end{align*}
In particular, we can compute:
\begin{align*}
M^k
&= (VDV^{-1})(VDV^{-1})\cdots(VDV^{-1})
\\&= VDV^{-1}VDV^{-1}\cdots V DV^{-1}
\\&= VDVD\cdots DV^{-1}
\\&= VD^{k}V^{-1}.
\end{align*}
Then we can write the matrix–exponential function as:
\begin{align*}
e^{Mt}
&= \sum_{k=0}^{\infty} \frac{M^k}{k!}t^k
\\[4mm]&=\sum_{k=0}^{\infty} \frac{VD^kV^{-1}}{k!}t^k
\\[4mm]&= V\left(\sum_{k=0}^{\infty} \frac{D^k}{k!}t^k\right)V^{-1}.
\end{align*}
Explicitly, the sum is:
\begin{align*}
\sum_{k=0}^{\infty} \frac{D^k}{k!}t^k
&= \sum_{k=0}^{\infty}
\frac{t^k}{k!}
\begin{pmatrix}
\lambda_1^k & 0 & \cdots & 0\\
0 & \lambda_2^k & \cdots & 0\\
0 & 0 & \ddots & 0\\
0 & 0 & \cdots & \lambda_n^k
\end{pmatrix}
\\[4mm]&= \sum_{k=0}^{\infty}
\begin{pmatrix}
\frac{\lambda_1^kt^k}{k!} & 0 & \cdots & 0\\
0 & \frac{\lambda_2^kt^k}{k!} & \cdots & 0\\
0 & 0 & \ddots & 0\\
0 & 0 & \cdots & \frac{\lambda_n^kt^k}{k!}
\end{pmatrix}
\\[4mm]&= \begin{pmatrix}
\sum_{k=0}^{\infty} \frac{\lambda_1^kt^k}{k!} & 0 & \cdots & 0\\
0 & \sum_{k=0}^{\infty} \frac{\lambda_2^kt^k}{k!} & \cdots & 0\\
0 & 0 & \ddots & 0\\
0 & 0 & \cdots & \sum_{k=0}^{\infty} \frac{\lambda_n^kt^k}{k!}
\end{pmatrix}
\\[4mm]&=
\begin{pmatrix}
e^{\lambda_1 t} & 0 & \cdots & 0\\
0 & e^{\lambda_2 t} & \cdots & 0\\
0 & 0 & \ddots & 0\\
0 & 0 & \cdots & e^{\lambda_n t}
\end{pmatrix}.
\end{align*}
Putting this all together, we see that:
\begin{align*}
e^{Mt}
=
V\begin{pmatrix}
e^{\lambda_1 t} & 0 & \cdots & 0\\
0 & e^{\lambda_2 t} & \cdots & 0\\
0 & 0 & \ddots & 0\\
0 & 0 & \cdots & e^{\lambda_n t}
\end{pmatrix}
V^{-1}.
\end{align*}
Now, we consider the system of differential equations with initial condition:
\begin{align*}
\boldsymbol{x}'(t)
= M\boldsymbol{x}(t),
\hspace{.5in}
\boldsymbol{x}(0) = \boldsymbol{c},
\end{align*}
where $$\boldsymbol{c}$$ is some constant vector. Then the solution to this initial value problem is:
\begin{align*}
\boldsymbol{x}(t)
= e^{Mt}\boldsymbol{c}.
\end{align*}
Indeed, we may compute:
\begin{align*}
\frac{d}{dt}e^{Mt}\boldsymbol{c}
&= \left(\sum_{k=0}^{\infty}\frac{d}{dt}\frac{(Mt)^k}{k!}\right)\boldsymbol{c}
\\[4mm]&= \left(\sum_{k=0}^{\infty}\frac{kM^kt^{k-1}}{k!}\right)\boldsymbol{c}
\\[4mm]&= \left(\sum_{k=1}^{\infty}\frac{M^kt^{k-1}}{(k-1)!}\right)\boldsymbol{c}
\\[4mm]&= \left(\sum_{k=0}^{\infty}\frac{M^{k+1}t^{k}}{k!}\right)\boldsymbol{c}
\\[4mm]&= M e^{Mt}\boldsymbol{c},
\end{align*}
as desired.

#### Examples

1. Find $$\exp{\begin{pmatrix}3 & -1\\4&-2\end{pmatrix}t}.$$

We've seen that the eigenvalues are $$-1$$ and $$2$$ with corresponding eigenvectors $$\begin{pmatrix}1\\4\end{pmatrix}$$ and $$\begin{pmatrix}1\\1\end{pmatrix}$$. In the previous video, we computed
$$\hspace{.5in} V =\begin{pmatrix}1 & 1\\ 4 & 1\end{pmatrix} \hspace{.5in}$$ and $$\hspace{.5in} V^{-1} = \begin{pmatrix}-1/3&1/3\\4/3&-1/3\end{pmatrix} \hspace{.5in}$$ and $$\hspace{.5in} D = \begin{pmatrix}-1&0\\0&2\end{pmatrix}$$

Then
\begin{align*}
e^{Mt}
&=\begin{pmatrix}1 & 1\\ 4 & 1\end{pmatrix}
\begin{pmatrix}e^{-t} & 0\\ 0 & e^{2t}\end{pmatrix}
\begin{pmatrix}-1/3&1/3\\4/3&-1/3\end{pmatrix}
\\[4mm]&=\frac{1}{3}\begin{pmatrix}1 & 1\\ 4 & 1\end{pmatrix}
\begin{pmatrix}-e^{-t} & e^{-t}\\4e^{2t} & -e^{2t}\end{pmatrix}
\\[4mm]&=\frac{1}{3}\begin{pmatrix}4e^{2t}-e^{-t} & -e^{2t} + e^{-t}\\
4e^{2t} - 4e^{-t} & -e^{2t} + 4e^{-t}\end{pmatrix}.
\end{align*}

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L7V5

2. Solve the system of differential equations:
\begin{align*}
x_1'(t) &= 3x_1(t) - x_2(t)\\
x_2'(t) &= 4x_1(t) -2x_2(t).
\end{align*}
with initial condition $$x_1(0) = c_1$$ and $$x_2(0) = c_2.$$

This can be written in matrix–vector form:
\begin{align*}
\begin{pmatrix}x_1'(t)\\x_2'(t)\end{pmatrix}
=
\begin{pmatrix}3&-1\\4&-2\end{pmatrix}
\begin{pmatrix}x_1(t)\\x_2(t)\end{pmatrix}.
\end{align*}
Then the solution is:
\begin{align*}
\begin{pmatrix}
x_1(t)\\x_2(t)
\end{pmatrix}
=
\left(\exp \left\{t \begin{pmatrix}3&-1\\4&-2\end{pmatrix}\right\}\right)
\begin{pmatrix}c_1\\c_2\end{pmatrix}.
\end{align*}
In the above, we saw that:
\begin{align*}
e^{Mt}
=\frac{1}{3}\begin{pmatrix}4e^{2t}-e^{-t} & -e^{2t} + e^{-t}\\
4e^{2t} - 4e^{-t} & -e^{2t} + 4e^{-t}\end{pmatrix}.
\end{align*}
And so, the solution to the initial value problem is:
\begin{align*}
\begin{pmatrix}
x_1(t)\\x_2(t)
\end{pmatrix}
&= \frac{1}{3}\begin{pmatrix}4e^{2t}-e^{-t} & -e^{2t} + e^{-t}\\
4e^{2t} - 4e^{-t} & -e^{2t} + 4e^{-t}\end{pmatrix}
\begin{pmatrix}c_1\\c_2\end{pmatrix}
\\[4mm]&=\frac{1}{3}\begin{pmatrix}(4c_1-c_2)e^{2t} + (-c_1 + c_2)e^{-t}\\
(4c_1 - c_2)e^{2t} + (-4c_1 + 4c_2)e^{-t}\end{pmatrix}.
\end{align*}
If, for example, $$c_1=1$$ and $$c_2=0$$ this is:
\begin{align*}
\begin{pmatrix}x_1(t)\\x_2(t)\end{pmatrix}
= \frac{1}{3}\begin{pmatrix}4e^{2t} -e^{-t}\\4e^{2t} -4e^{-t}\end{pmatrix}.
\end{align*}

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L7V6

3. Solve the system of differential equations:
\begin{align*}
x_1'(t) &= 3x_1(t) - x_2(t) - 2x_2\\
x_2'(t) &= 2x_2(t)\\
x_3'(t) &= x_1(t) - x_2(t)
\end{align*}

This can be written in matrix–vector form:
\begin{align*}
\begin{pmatrix}x_1'(t)\\x_2'(t)\\x_3'(t)\end{pmatrix}
=
\begin{pmatrix}3&-1&-2\\0 & 2 & 0\\1&-2&0\end{pmatrix}
\begin{pmatrix}x_1(t)\\x_2(t)\\x_3(t)\end{pmatrix}.
\end{align*}
Then the solution is:
\begin{align*}
\begin{pmatrix}
x_1(t)\\x_2(t)\\x_3(t)
\end{pmatrix}
=
\left(\exp \left\{t \begin{pmatrix}3&-1&-2\\0 & 2 & 0\\1&-2&0\end{pmatrix}\right\}\right)
\begin{pmatrix}c_1\\c_2\\c_3\end{pmatrix}.
\end{align*}
First, we need to find the eigenvalues and eigenvectors:
\begin{align*}
\det (M-\lambda I)&= det \begin{pmatrix}3-\lambda&-1&-2\\0 & 2-\lambda & 0\\1&-2&-\lambda\end{pmatrix}\\[4mm]
&=(3-\lambda)\lambda(\lambda-2)-2(\lambda-2)\\
&=(\lambda-2)(\lambda(3-\lambda)-2)\\
&=-(\lambda-2)(\lambda^2-3\lambda+2)\\
&=-(\lambda-2)^2(\lambda-1)
\end{align*}
Therefore, the eigenvalues are $$\lambda_1=1$$, $$\lambda_2=2$$, and $$\lambda_3=3$$. Now we calculate the eigenvectors. For $$\lambda_1=1$$, we row reduce $$M-1I$$ to get
\begin{align*}
\begin{pmatrix}
2 & -1 & -2\\
0 & 1 & 0\\
1 & -1 & -1\\
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 0 & -1\\
0 & 1 & 0\\
0 & 0 & 0\\
\end{pmatrix}
\end{align*}
Therefore, the eigenvectors for $$\lambda_1=1$$ are of the form $$t\begin{pmatrix} 1\\ 0 \\1 \end{pmatrix}$$ so a basis for the eigenspace is  $$\begin{pmatrix} 1\\0\\1\end{pmatrix}$$. So we will use the eigenvector $$\boldsymbol{v}_1= \begin{pmatrix} 1\\0\\1\end{pmatrix}.$$

For $$\lambda_2=2$$, we row reduce $$M-2I$$ to get
\begin{align*}
\begin{pmatrix}
1 & -1 & -2\\
0 & 0 & 0\\
1 & -1 & -2\\
\end{pmatrix}
\sim
\begin{pmatrix}
1 & -1 & -2\\
0 & 0 & 0\\
0 & 0 & 0\\
\end{pmatrix}
\end{align*}
Therefore, the eigenvectors for $$\lambda_2=2$$ are of the form $$s\begin{pmatrix}1\\1\\0\end{pmatrix}+t\begin{pmatrix} 2\\ 0 \\1 \end{pmatrix}$$. So we will use the eigenvectors $$\boldsymbol{v}_2= \begin{pmatrix} 1\\1\\0\end{pmatrix}$$ and $$\boldsymbol{v}_3= \begin{pmatrix} 2\\0\\1\end{pmatrix}.$$

Then $$M=VDV^{-1}$$ so we need to calculate $$V^{-1}$$ for the matrix $$V=(\boldsymbol{v}_1, \boldsymbol{v}_2, \boldsymbol{v}_3)$$. We rowreduce the augmented matrix $$(M | I)$$ to get
\begin{align*}
\left(\begin{array}{ccc|ccc}
1 & 2 & 2 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0\\
1 & 0 & 1 & 0 & 0 & 1
\end{array}\right)
\sim
\left(\begin{array}{ccc|ccc}
1 & 0 & 0 & -1 & 1 & 2\\
0 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 1 & -1 & -1
\end{array}\right)
\end{align*}
The right–hand side of this is $$V^{-1}$$ so we can write $$M=VDV^{-1}$$ as
\begin{align*}
M=
\begin{pmatrix}
1 & 1 & 2\\
0 & 1 & 0 \\
1 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
​1 & 0 & 0\\
0 & 2 & 0 \\
0 & 0 & 2
\end{pmatrix}
\begin{pmatrix}
​-1 & 1 & 2\\
0 & 1 & 0 \\
1 & -1 & -1
\end{pmatrix}
\end{align*}
Now we can compute
\begin{align*}
e^{Mt}&=
\begin{pmatrix}
1 & 1 & 2\\
0 & 1 & 0 \\
1 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
e^t & 0 & 0\\
0 & e^{2t} & 0 \\
0 & 0 & e^{2t}
\end{pmatrix}
\begin{pmatrix}
​-1 & 1 & 2\\
0 & 1 & 0 \\
1 & -1 & -1
\end{pmatrix}
\\[4mm]&=
\begin{pmatrix}
2e^{2t}-e^t & -e^{2t}+e^t & -2e^{2t}+2e^t\\
0 & e^{2t} & 0\\
e^{2t}-e^t & -e^{2t}+e^t & -e^{2t}+2e^t
\end{pmatrix}
\end{align*}
The solution to the equation is
\begin{align*}
e^{Mt}\begin{pmatrix}c_1\\c_2\\c_3\end{pmatrix}
&=\begin{pmatrix}
2e^{2t}-e^t & -e^{2t}+e^t & -2e^{2t}+2e^t\\
0 & e^{2t} & 0\\
e^{2t}-e^t & -e^{2t}+e^t & -e^{2t}+2e^t
\end{pmatrix}
\begin{pmatrix}c_1\\c_2\\c_3\end{pmatrix}
\\[4mm]&=
\begin{pmatrix}
(2c_1-c_2-2c_3)e^{2t}+(-c_1+c_2+2c_3)e^t\\
c_2e^{2t}\\
(c_1-c_2-c_3)e^{2t}+(-c_1+c_2+2c_3)e^{2t}
\end{pmatrix}
\end{align*}

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L7V7

Exercises

1. Compute $$\exp{t\begin{pmatrix} 5 & -1\\ 3 & 1 \end{pmatrix}} .$$

In a previous video, we found:
\begin{align*}
\lambda_1 = 2, \hspace{.25in}\boldsymbol{v}_1 = \begin{pmatrix}1\\3\end{pmatrix}, \hspace{.5in}
\lambda_2 = 4, \hspace{.25in}\boldsymbol{v}_2 = \begin{pmatrix}1\\1\end{pmatrix}.
\end{align*}
So then:
\begin{align*}
\exp{t\begin{pmatrix}
5 & -1\\
3 & 1
\end{pmatrix}}
&=\begin{pmatrix}1 & 1\\ 3 & 1\end{pmatrix}
\begin{pmatrix}e^{2t} & 0\\0 & e^{4t}\end{pmatrix}
\begin{pmatrix}1 & 1\\ 3 & 1\end{pmatrix}^{-1}
\\[4mm]&=-\frac{1}{2}\begin{pmatrix}1 & 1\\ 3 & 1\end{pmatrix}
\begin{pmatrix}e^{2t} & 0\\0 & e^{4t}\end{pmatrix}
\begin{pmatrix}1 & -1\\ -3 & 1\end{pmatrix}
\\[4mm]&= -\frac{1}{2}\begin{pmatrix}1 & 1\\ 3 & 1\end{pmatrix}
\begin{pmatrix}
e^{2t} & -e^{2t}\\-3e^{4t} & e^{4t}
\end{pmatrix}
\\[4mm]&=-\frac{1}{2}
\begin{pmatrix}
e^{2t} - 3e^{4t} & -e^{2t} + e^{4t}\\
3e^{2t} - 3e^{4t} & -3e^{2t} + e^{4t}
\end{pmatrix}.
\end{align*}

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L7E1

2. Solve $$\boldsymbol{x}'(t) = \begin{pmatrix} 5 & -1\\ 3 & 1 \end{pmatrix}\boldsymbol{x}(t),$$ with $$\boldsymbol{x}(0)=\begin{pmatrix}1\\0\end{pmatrix}.$$

The solution is:
\begin{align*}
\boldsymbol{x}(t) = e^{tM}\boldsymbol{x}(0).
\end{align*}
We compute $$e^{tM}$$ for this matrix above and so the solution is:
\begin{align*}
\boldsymbol{x}(t)
&= e^{tM}\boldsymbol{x}(0)\\
&= -\frac{1}{2}
\begin{pmatrix}
e^{2t} - 3e^{4t} & -e^{2t} + e^{4t}\\
3e^{2t} - 3e^{4t} & -3e^{2t} + e^{4t}
\end{pmatrix}
\begin{pmatrix}1\\0\end{pmatrix}
\\[4mm]&= -\frac{1}{2}\begin{pmatrix}
e^{2t} - 3e^{4t}\\3e^{2t} - 3e^{4t}
\end{pmatrix}.
\end{align*}

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L7E2

3. Compute $$\exp{t \begin{pmatrix} -2 & 1\\ 1 & -2 \end{pmatrix}}$$

We have already found the eigenvalues and eigenvectors in a previous lecture. They are:
\begin{align*}
\lambda_1 = -3, \hspace{.25in}\boldsymbol{v}_1 = \begin{pmatrix}1\\-1\end{pmatrix}, \hspace{.5in}
\lambda_2 = -1, \hspace{.25in}\boldsymbol{v}_2 = \begin{pmatrix}1\\1\end{pmatrix}.
\end{align*}
So then:
\begin{align*}
e^{tM}
&= \begin{pmatrix}1 & 1\\ -1 & 1\end{pmatrix}
\begin{pmatrix}e^{-3t} & 0\\ 0 & e^{-t}\end{pmatrix}
\begin{pmatrix}1 & 1\\ -1 & 1\end{pmatrix}^{-1}
\\[4mm]&=\frac{1}{2}\begin{pmatrix}1 & 1\\ -1 & 1\end{pmatrix}
\begin{pmatrix}e^{-3t} & 0\\ 0 & e^{-t}\end{pmatrix}
\begin{pmatrix}1 & -1\\ 1 & 1\end{pmatrix}
\\[4mm]&=\frac{1}{2}\begin{pmatrix}1 & 1\\ -1 & 1\end{pmatrix}
\begin{pmatrix}
e^{-3t} & -e^{-3t}\\
e^{-t} & e^{-t}
\end{pmatrix}
\\[4mm]&=\frac{1}{2}
\begin{pmatrix}
e^{-3t} + e^{-t} & -e^{-3t} + e^{-t}\\
-e^{-3t} + e^{-t} & e^{-3t} + e^{-t}
\end{pmatrix}.
\end{align*}

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L7E3

4. Solve $$\boldsymbol{x}'(t) = \begin{pmatrix} -2 & 1\\ 1 & -2 \end{pmatrix}\boldsymbol{x}(t)$$, with $$\boldsymbol{x}(0)=\begin{pmatrix}1\\1\end{pmatrix}.$$

We found the matrix exponential above. So the solution to this initial value problem is:
\begin{align*}
\boldsymbol{x}(t)
= e^{tM}\begin{pmatrix}1\\1\end{pmatrix}
= \frac{1}{2}
\begin{pmatrix}
e^{-3t} + e^{-t} & -e^{-3t} + e^{-t}\\
-e^{-3t} + e^{-t} & e^{-3t} + e^{-t}
\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix}.
\end{align*}

At this point, you can carry out this computation (as was done in the video) or use this interesting fact. Note that $$\boldsymbol{x}(0)=\begin{pmatrix}1\\1\end{pmatrix}$$ is an eigenvector of the matrix associated to the eigenvalue $$-1$$. We may compute as follows:
\begin{align*}
e^{tM}\boldsymbol{x}(0)
&=\sum_{k=0}^{\infty}\frac{t^kM^k}{k!}\boldsymbol{x}(0)
\\[2mm]&=\sum_{k=0}^{\infty}\frac{t^k \lambda^k\boldsymbol{x}(0)}{k!}
\\[2mm]&=e^{\lambda t}\boldsymbol{x_0}
\\[2mm]&=e^{-t}\begin{pmatrix}1\\1\end{pmatrix}.
\end{align*}
Above, we used the fact that $$M^k\boldsymbol{x}_0 = \lambda^k\boldsymbol{x}(0).$$

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L7E4

5. Compute $$\exp{t \begin{pmatrix} 3 & -2\\ 4 & -1 \end{pmatrix}}$$.

We have already found the eigenvalues and eigenvectors in a previous lecture. They are:
\begin{align*}
\lambda_1 = 1+2i, \hspace{.25in}\boldsymbol{v}_1 = \begin{pmatrix}1\\1-i\end{pmatrix}, \hspace{.5in}
\lambda_2 = 1-2i, \hspace{.25in}\boldsymbol{v}_2 = \begin{pmatrix}1\\1+i\end{pmatrix}.
\end{align*}
So the matrix exponential is:
\begin{align*}
e^{tM}
&= \begin{pmatrix} 1 & 1\\ 1-i & 1+i\end{pmatrix}
\begin{pmatrix}e^{(1+2i)t} & 0 \\ 0 & e^{(1-2i)t}\end{pmatrix}
\begin{pmatrix} 1 & 1\\ 1-i & 1+i\end{pmatrix}^{-1}
\\[4mm]&=\frac{e^{t}}{2}\begin{pmatrix} 1 & 1\\ 1-i & 1+i\end{pmatrix}
\begin{pmatrix}e^{2it} & 0 \\ 0 & e^{-2it}\end{pmatrix}
\begin{pmatrix}1-i & i\\ 1+i & -i\end{pmatrix}
\\[4mm]&=\frac{e^t}{2}\begin{pmatrix} 1 & 1\\ 1-i & 1+i\end{pmatrix}
\begin{pmatrix}
(1-i)e^{2it} & ie^{2it}\\
(1+i)e^{-2it} & -ie^{-2it}
\end{pmatrix}
\\[4mm]&=\frac{e^t}{2}
\begin{pmatrix}
(1-i)e^{2it} + (1+i)e^{-2it} & ie^{2it}-ie^{-2it}\\
(1-i)^2e^{2it} + (1+i)^2e^{-2it} & (1+i)e^{2it} & (1-i)e^{-2it}
\end{pmatrix}.
\end{align*}
Observe that each entry is of the form $$z + \bar{z}$$ where $$\bar{z}$$ is the complex conjugate. That is, if $$z = a+bi$$, then $$\bar{z} = a-bi$$. Note that $$z + \bar{z} = 2 \textnormal{Re}z$$. That is: $$(a+ib) + (a-ib) = 2a$$. So we have to compute the real and imaginary of each entry. To do this, we use Euler's identity: $$e^{it} = \cos t + i \sin t$$.
\begin{align*}
(1-i)e^{2it} = (1-i)(\cos 2t + i \sin 2t)
&\Rightarrow \textnormal{Re} (1-i)e^{2it} = \cos 2t + \sin 2t\\
ie^{2it} = i(\cos 2t + i\sin 2t)
&\Rightarrow \textnormal{Re} ie^{2it} = -\sin 2t\\
(1-i)^2e^{2it} = (1-i)^2(\cos 2t + i \sin 2t) = -2i(\cos 2t + i\sin 2t)
&\Rightarrow \textnormal{Re} ie^{2it} = 2\sin 2t\\
(1+i)e^{2it} = (1+i)(\cos 2t + i \sin 2t)
&\Rightarrow \textnormal{Re}(1+i)e^{2it} = \cos 2t - \sin 2t.
\end{align*}
So then:
\begin{align*}
\frac{e^t}{2}
\begin{pmatrix}
2(\cos 2t + \sin 2t) & 2(-\sin 2t)\\
2(2\sin 2t) & 2(\cos 2t - \sin 2t)
\end{pmatrix}
=e^{t}\begin{pmatrix}
\cos 2t + \sin 2t & -\sin 2t\\
2\sin 2t & \cos 2t - \sin 2t
\end{pmatrix}.
\end{align*}
Note: we had found the "correct" matrix exponential even in the matrix that had "$$i$$'s" in it. However, since the given matrix has only real entries, and in the series definition of $$e^{tM}$$ there are no non–real numbers, it makes sense that the matrix exponential would have only real entries. So we use some basic facts about complex numbers to show that it does only contain real entries and we were able to compute them as well.

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Linear Algebra for Math 308: L7E5

6. Solve $$\boldsymbol{x}'(t) = \begin{pmatrix} 3 & -2\\ 4 & -1 \end{pmatrix}\boldsymbol{x}(t)$$, with $$\boldsymbol{x}(0)=\begin{pmatrix}1\\1\end{pmatrix}.$$

We found the matrix exponential in the previous exercise. So the solution to the initial value problem is:
\begin{align*}
e^{tM}\begin{pmatrix}1\\1\end{pmatrix}
&= e^{t}\begin{pmatrix}
\cos 2t + \sin 2t & -\sin 2t\\
2\sin 2t & \cos 2t - \sin 2t
\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix}
\\&=e^{t}\begin{pmatrix}
\cos 2t\\ \sin 2t + \cos 2t
\end{pmatrix}.
\end{align*}

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Linear Algebra for Math 308: L5V6

7. Compute $$\exp{t\begin{pmatrix} 1 & 0 & 0\\ -3 & -2 &3\\ 0 & 0 & 1 \end{pmatrix}}$$

We have already found the eigenvalues and eigenvectors:
$$\hspace{.25in} \lambda_1 = -2, \hspace{.25in} \boldsymbol{v}_1 = \begin{pmatrix}0\\1\\0\end{pmatrix}, \hspace{.5in} \lambda_{2,3} = 1, \hspace{.25in}$$$$\boldsymbol{v}_2 = \begin{pmatrix}-1\\1\\0\end{pmatrix}, \hspace{.25in} \boldsymbol{v}_3 = \begin{pmatrix}1\\0\\1\end{pmatrix}.$$
So then:
\begin{align*}
e^{tM}
= \begin{pmatrix}
0 & -1 & 1\\
1 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
e^{-2t} & 0& 0\\
0 & e^{t} & 0\\
0 & 0 & e^{t}
\end{pmatrix}
\begin{pmatrix}
0 & -1 & 1\\
1 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}^{-1}.
\end{align*}
Now we compute the inverse:
\begin{align*}
\left(\begin{array}{ccc|ccc}
0 & -1 & 1 & 1 & 0 & 0\\
1 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 &1
\end{array}\right)
\xrightarrow{R1 \leftrightarrow R2}
\left(\begin{array}{ccc|ccc}
1 & 1 & 0 & 0 & 1 & 0\\
0 & -1 & 1 & 1 & 0 & 0\\
0 & 0 & 1 & 0 & 0 &1
\end{array}\right)
\\[4mm]\xrightarrow{R2 = -(R2 - R3)}
\left(\begin{array}{ccc|ccc}
1 & 1 & 0 & 0 & 1 & 0\\
0 & 1 & 0 & -1 & 0 & 1\\
0 & 0 & 1 & 0 & 0 &1
\end{array}\right)
\\[4mm]\xrightarrow{R1 = R1 - R2}
\left(\begin{array}{ccc|ccc}
1 & 0 & 0 & 1 & 1 & -1\\
0 & 1 & 0 & -1 & 0 & 1\\
0 & 0 & 1 & 0 & 0 &1
\end{array}\right)
\end{align*}
So then:
\begin{align*}
e^{tM}
&=
\begin{pmatrix}
0 & -1 & 1\\
1 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
e^{-2t} & 0& 0\\
0 & e^{t} & 0\\
0 & 0 & e^{t}
\end{pmatrix}
\begin{pmatrix}
1 & 1 & -1\\
-1 & 0 & 1\\
0 & 0 & 1
\end{pmatrix}
\\[4mm]&=\begin{pmatrix}
0 & -1 & 1\\
1 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
e^{-2t} & e^{-2t} & -e^{-2t}\\
-e^{t} & 0 & e^{t}\\
0 & 0 & e^{t}
\end{pmatrix}
\\[4mm]&= \begin{pmatrix}
e^{t} & 0 & 0\\
e^{-2t} - e^{t} & e^{-2t} & -e^{-2t} + e^{t}\\
0 & 0 & e^{t}
\end{pmatrix}.
\end{align*}

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Linear Algebra for Math 308: L7E7

8. Solve $$\boldsymbol{x}'(t) = \begin{pmatrix} 1 & 0 & 0\\ -3 & -2 &3\\ 0 & 0 & 1 \end{pmatrix}\boldsymbol{x}(t)$$ with $$\boldsymbol{x}(0) = \begin{pmatrix}-1\\0\\1\end{pmatrix}$$

We found $$e^{tM}$$ above so we use that:
\begin{align*}
\boldsymbol{x}(t)
&= e^{tM}\boldsymbol{x}(0)
\\[4mm]&= \begin{pmatrix}
e^{t} & 0 & 0\\
e^{-2t} - e^{t} & e^{-2t} & -e^{-2t} + e^{t}\\
0 & 0 & e^{t}
\end{pmatrix}
\begin{pmatrix}-1\\0\\1\end{pmatrix}
\\[4mm]&= \begin{pmatrix}
-e^{t}\\
-2e^{-2t} + 2e^{t}\\
e^t
\end{pmatrix}.
\end{align*}

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Linear Algebra for Math 308: L7E8