# Lecture 8: 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.
• Thisis the last section so you can return to the main page when you have finished the material.

### Concepts

Systems of Differential Equations

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#### Video Notes

Sometimes there are two quantities we are interested in and their derivatives depend on each other. (For example, predator-prey system, two tanks of water with salt, etc). Sometimes the equations can be written in the form
\begin{align*}
x_1'(t) &= ax_1(t) + bx_2(t)\\
x_2'(t) &= cx_1(t) + dx_2(t),
\end{align*}
where $$a,b,c,d$$ are constants. An equation of this form is called an first order system of linear equations. Using matrix–vector notation we can write it as:
\begin{align*}
\begin{pmatrix}x_1'(t)\\x_2'(t)\end{pmatrix}
=\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}
\begin{pmatrix}x_1(t)\\x_2(t)\end{pmatrix}.
\end{align*}

Recall that the definition of matrix–vector multiplication is:
\begin{align*}
\begin{pmatrix}a & b\\ c & d\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}
:= x_1\begin{pmatrix}a\\c\end{pmatrix} + x_2\begin{pmatrix}b\\d\end{pmatrix}
= \begin{pmatrix}ax_1 + bx_2\\cx_1 + dx_2\end{pmatrix}.
\end{align*}

If the matrix–vector notation stuff was just another way to write the system, it wouldn't be helpful. But, it allows us to use tools from linear algebra. Let's look at a specific example and see how we can use linear algebra.

To illustrate the basic technique, we will solve
\begin{align*}
\begin{pmatrix}x_1'(t)\\x_2'(t)\end{pmatrix}
=\begin{pmatrix}1&2\\2&1\end{pmatrix}\begin{pmatrix}x_1(t)\\x_2(t)\end{pmatrix}.
\end{align*}
You may verify for yourself that the eigenvalues and eigenvectors are:
$$\hspace{1in} \lambda_1 = -1, \hspace{.25in} \boldsymbol{v}_1 = \begin{pmatrix}-1\\1\end{pmatrix}, \hspace{1in}$$$$\lambda_2 = 3, \hspace{.25in} \boldsymbol{v}_2 = \begin{pmatrix}1\\1\end{pmatrix}.$$

A solution to this ODE is a vector, $$\begin{pmatrix}x_1(t)\\x_2(t)\end{pmatrix}$$. Since the two eigenvectors form a basis for $$\mathbb{R}^2$$, then for each fixed $$t$$ there are scalars $$c_1(t)$$ and $$c_2(t)$$ such that:
\begin{align*}
\begin{pmatrix}
x_1(t)\\x_2(t)
\end{pmatrix}
= c_1(t)\begin{pmatrix}-1\\1\end{pmatrix}
+ c_2(t)\begin{pmatrix}1\\1\end{pmatrix}.
\end{align*}
If we plug this into the ODE we get:
\begin{align*}
\left(c_1(t)\begin{pmatrix}-1\\1\end{pmatrix}
+ c_2(t)\begin{pmatrix}1\\1\end{pmatrix}\right)'
&= c_1'(t)\begin{pmatrix}-1\\1\end{pmatrix}
+ c_2'(t)\begin{pmatrix}1\\1\end{pmatrix}
\\&= \begin{pmatrix}1&2\\2&1\end{pmatrix}
\left(c_1(t)\begin{pmatrix}-1\\1\end{pmatrix}
+ c_2(t)\begin{pmatrix}1\\1\end{pmatrix}\right)
\\&=c_1(t)\begin{pmatrix}1&2\\2&1\end{pmatrix}\begin{pmatrix}-1\\1\end{pmatrix}
+c_2(t)\begin{pmatrix}1&2\\2&1\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix}
\\& -c_1(t)\begin{pmatrix}-1\\1\end{pmatrix} + 3c_2(t)\begin{pmatrix}1\\1\end{pmatrix}.
\end{align*}
This can be re–written as:
\begin{align*}
(c_1'(t) + c_1(t))\begin{pmatrix}-1\\1\end{pmatrix}
+ (c_2'(t) - 3c_2(t))\begin{pmatrix}1\\1\end{pmatrix}
= \begin{pmatrix}0\\0\end{pmatrix}.
\end{align*}
These two vectors are linearly independent and so for every value of $$t$$ we deduce:
\begin{align*}
c_1'(t) + c_1(t) & = 0\\
c_2'(t) - 3c_2(t) & = 0
\end{align*}
whence:
\begin{align*}
c_1(t) &= a_1 e^{-t}\\
c_2(t) &= a_2 e^{3t}.
\end{align*}
And so the solution to the ODE is:
\begin{align*}
\begin{pmatrix}
x_1(t)\\x_t(2)
\end{pmatrix}
= a_1e^{-t}\begin{pmatrix}-1\\1\end{pmatrix}
+ a_2e^{3t}\begin{pmatrix}1\\1\end{pmatrix}.
\end{align*}
The coefficients $$a_1$$ and $$a_2$$ are arbitrary and we can solve for them if we had an initial condition like:
\begin{align*}
\begin{pmatrix}x_1(0)\\x_2(0)\end{pmatrix} = \begin{pmatrix}a\\b\end{pmatrix}.
\end{align*}

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#### Video Notes

In general, assume we have an $$n\times n$$ system:
\begin{align*}
\boldsymbol{x}' = M\boldsymbol{x}
\end{align*}
and that the matrix $$M$$ has $$n$$ linearly independent eigenvectors. Then the solution can be written as:
\begin{align*}
\boldsymbol{x}(t) = \sum_{k=1}^{n}c_1(t)\boldsymbol{v}_k,
\end{align*}
for to be determined scalar–valued functions $$c_1,\ldots, c_n$$. If we plug this into the ODE we get:
\begin{align*}
\sum_{k=1}^{n}c_k'(t)\boldsymbol{v}_k
=M\sum_{k=1}^{n}c_k(t)\boldsymbol{v}_k
=\sum_{k=1}^{n}c_k(t)M\boldsymbol{v}_k
=\sum_{k=1}^{n}c_k(t)\lambda_k \boldsymbol{v}_k.
\end{align*}
Since the eigenvectors are linearly independent, this implies that the corresponding coefficients are equal:
\begin{align*}
c_1'(t) &= \lambda_1 c_1(t)\\
&\vdots\\
c_n'(t) &= \lambda_n c_n(t).
\end{align*}
And all of these are scalar–valued equations we know how to solve, and the general solutions are
\begin{align*}
c_1(t) &= A_1e^{\lambda_1 t}\\
&\vdots\\
c_n(t) &= A_ne^{\lambda_n t}.
\end{align*}
where the $$A_k$$ are arbitrary constants. Therefore, the general solution to the differential equation is
\begin{align*}
\boldsymbol{x}(t) = \sum_{k=1}^{n}A_ke^{\lambda_k t}\boldsymbol{v}_k
\end{align*}

#### Examples

Find the general solution to the IVP:
\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},
\hspace{.5in}
\begin{pmatrix}x_1(0)\\x_2(0)\end{pmatrix}
=\begin{pmatrix}1\\0\end{pmatrix}.
\end{align*}

The eigenvalues and eigenvectors are:
$$\hspace{.5in} \lambda_1 = -1, \hspace{.25in} \boldsymbol{v}_1 = \begin{pmatrix}1\\4\end{pmatrix} \hspace{.5in}$$$$\lambda_2 = 2, \hspace{.25in} \boldsymbol{v}_2 = \begin{pmatrix}1\\1\end{pmatrix}.$$
Then we get the system:
\begin{align*}
c_1' &= -1c_1\\
c_2' &= 2c_2.
\end{align*}
And so:
\begin{align*}
c_1(t) &= a_1e^{-t}\\
c_2(t) &= a_2e^{2t}.
\end{align*}
So the solution is:
\begin{align*}
\begin{pmatrix}x_1(t)\\x_2(t)\end{pmatrix}
=a_1e^{-t}\begin{pmatrix}1\\4\end{pmatrix}
+a_2e^{2t}\begin{pmatrix}1\\1\end{pmatrix}.
\end{align*}
We use the initial condition to find the coefficients:
\begin{align*}
\begin{pmatrix}1\\0\end{pmatrix}
=\begin{pmatrix}x_1(0)\\x_2(0)\end{pmatrix}
=a_1\begin{pmatrix}1\\4\end{pmatrix}
+a_2\begin{pmatrix}1\\1\end{pmatrix}.
\end{align*}
This gives $$a_1 = -\frac{1}{3}$$ and $$a_2 = \frac{4}{3}$$ so the solution to the IVP is:
\begin{align*}
\begin{pmatrix}x_1(t)\\x_2(t)\end{pmatrix}
=-\frac{1}{3}e^{-t}\begin{pmatrix}1\\4\end{pmatrix}
+\frac{4}{3}e^{2t}\begin{pmatrix}1\\1\end{pmatrix}.
\end{align*}

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L8V3

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#### Video Notes

When there are $$n$$ linearly independent eigenvectors, the solution given above is always correct. However, if the eigenvalues are complex (and the entries of the matrix are real), then the eigenvectors will have complex entries. If the ODE is modelling something like a predator–prey system, then having non–real entries in the reported solution makes it hard to interpret. So in the case of complex eigenvalues, we do something slightly different. We illustrate the method with the following example:

Find the general solution to the ODE:
\begin{align*}
\begin{pmatrix}
x_1(t)\\x_2(t)
\end{pmatrix}'
=\begin{pmatrix}-.5&1\\-1&-.5\end{pmatrix}\begin{pmatrix}x_1(t)\\x_2(t)\end{pmatrix}.
\end{align*}
Given:
\begin{align*}
\begin{pmatrix}-.5&1\\-1&-.5\end{pmatrix}
\begin{pmatrix}1\\i\end{pmatrix} &= (-\frac{1}{2}+i)\begin{pmatrix}1\\i\end{pmatrix}\\
\begin{pmatrix}-.5&1\\-1&-.5\end{pmatrix}
\begin{pmatrix}1\\-i\end{pmatrix} &= (-\frac{1}{2}-i)\begin{pmatrix}1\\-i\end{pmatrix}.
\end{align*}
As you may verify, the eigenvalues and eigenvectors are:
$$\hspace{.5in} \lambda_1 = -\frac{1}{2} + i, \hspace{.25in} \boldsymbol{v}_1 = \begin{pmatrix}1\\i\end{pmatrix} \hspace{.5in}$$$$\lambda_2 = -\frac{1}{2} - i, \hspace{.25in} \boldsymbol{v}_2 = \begin{pmatrix}1\\-i\end{pmatrix}.$$
And so the solution as given above is:
\begin{align*}
\begin{pmatrix}x_1(t)\\x_2(t)\end{pmatrix}
&=a_1e^{-\frac{1}{2}t + it}\begin{pmatrix}1\\i\end{pmatrix}
+a_2e^{-\frac{1}{2}t - it}\begin{pmatrix}1\\-i\end{pmatrix}
\\[4mm]&=e^{-\frac{1}{2}t}
\left(a_1e^{ it}\begin{pmatrix}1\\i\end{pmatrix}
+a_2e^{- it}\begin{pmatrix}1\\-i\end{pmatrix}\right).
\end{align*}
Using Euler's formula: $$e^{-it} = \cos t + i \sin t$$, the part in parentheses above can be written as the following (note that $$\cos(-t)+i\sin(-t)=\cos(t)-i\sin(t)$$):
\begin{align*}
a_1(\cos t + i \sin t)\begin{pmatrix}1\\i\end{pmatrix}
+ a_2 (\cos t - i \sin t)\begin{pmatrix}1\\-i\end{pmatrix}.
\end{align*}
Splitting the vectors into real and imaginary parts, we get
\begin{align*}
a_1\left[
\begin{pmatrix} \cos t \\ -\sin t\end{pmatrix}
+i \begin{pmatrix} \sin t \\ \cos t\end{pmatrix}
\right]
+a_2\left[
\begin{pmatrix} \cos t \\ -\sin t\end{pmatrix}
-i\begin{pmatrix} \sin t\\ \cos t\end{pmatrix}
\right]
\end{align*}
And so the solution can be rewritten as:
\begin{align*}
\begin{pmatrix}x_1(t)\\x_2(t)\end{pmatrix} &=
(a_1+a_2)e^{-\frac{1}{2}t} \begin{pmatrix} \cos t \\ -\sin t\end{pmatrix}
+i (a_1-a_2) e^{-\frac{1}{2}t}\begin{pmatrix} \sin t \\ \cos t\end{pmatrix}\\
&=
b_1e^{-\frac{1}{2}t} \begin{pmatrix} \cos t \\ -\sin t\end{pmatrix}
+b_2 e^{-\frac{1}{2}t}\begin{pmatrix} \sin t \\ \cos t\end{pmatrix}
\end{align*}

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#### Video Notes

Let's assume we have the system of differential equations:
\begin{align*}
\boldsymbol{x}' = M\boldsymbol{x}
\end{align*}
where $$M$$ is a $$2\times 2$$ matrix and $$\boldsymbol{x}$$ is a two–dimensional vector. Let's also assume we have the eigenvalues and vectors:

$$\hspace{.25in} \lambda_1 = a + ib, \hspace{.25in} \boldsymbol{v}_1 = \boldsymbol{u} + i\boldsymbol{w}, \hspace{.5in} \lambda_2 = a - ib, \hspace{.25in}$$$$\boldsymbol{v}_2 = \boldsymbol{u} - i\boldsymbol{w},$$

where $$a$$ and $$b$$ are real and $$\boldsymbol{u}$$ and $$\boldsymbol{w}$$ are in $$\mathbb{R}^2$$. The theory we have learned so far says that the general solution is:
\begin{align*}
\boldsymbol{x}(t)
= A_1e^{(a+ib)t}\boldsymbol{v}_1 + A_2e^{(a-ib)t}\boldsymbol{v}_2.
\end{align*}
Using Euler's Formula and the fact that $$\cos(-bt) = \cos bt$$ and $$\sin(-bt)=-\sin bt$$, we can write this as:
\begin{align*}
A_1e^{(a+ib)t}\boldsymbol{v}_1 + A_2e^{(a-ib)t}\boldsymbol{v}_2
&= A_1e^{at}\left(\cos bt + i\sin bt\right)(\boldsymbol{u} + i \boldsymbol{w})
+ A_2e^{at}\left(\cos bt - i\sin bt\right)(\boldsymbol{u} - i \boldsymbol{w})
\\&=
(A_1 + A_2)e^{at}(\cos bt \boldsymbol{u}-\sin bt \boldsymbol{w}) +
(A_1 - A_2)ie^{at}(\cos bt \boldsymbol{w} + \sin bt \boldsymbol{u})
\\&= B_1e^{at}(\cos bt \boldsymbol{u}-\sin bt \boldsymbol{w})
+ B_2e^{at}(\cos bt \boldsymbol{w} + \sin bt \boldsymbol{u}),
\end{align*}
where $$B_1 = A_1 + A_2$$ and $$B_2 = i(A_1 - A_2)$$. In other words, we can report the general solution in terms of real vectors and real numbers (and we allow the numbers $$B_1$$ and $$B_2$$ to be complex). This is a better way to report the answer for many applications and this is what we require you to do in Math 308.

#### Example

Solve
\begin{align*}
\boldsymbol{x}'(t) = \begin{pmatrix}2 & 1\\ 0 & 2\end{pmatrix}\boldsymbol{x}(t).
\end{align*}

Note that there is only one eigenvalue: $$\lambda = 2$$ and one linearly independent eigenvector $$\boldsymbol{v} =\begin{pmatrix}1\\0\end{pmatrix}$$. if we solve the equation $$(M-\lambda I)\boldsymbol{w} = \boldsymbol{v}$$ to find the generalized eigenvector, we get $$\boldsymbol{w} = \begin{pmatrix}1\\0\end{pmatrix} + t\boldsymbol{v}$$. That is, we may take $$\boldsymbol{w} = \begin{pmatrix}1\\0\end{pmatrix}$$. The vectors $$\boldsymbol{v}$$ and $$\boldsymbol{w}$$ are linearly independent, so they form a basis for $$\mathbb{R}^2$$. I can proceed as I did in previous examples and write the solution vector as:
\begin{align*}
\boldsymbol{x}(t)
= c_1(t)\boldsymbol{v} + c_2(t)\boldsymbol{w}.
\end{align*}
If I plug this into the ODE I get:
\begin{align*}
c_1'(t)\begin{pmatrix}1\\0\end{pmatrix} + c_2'(t)\begin{pmatrix}0\\1\end{pmatrix}
&=c_1(t)M\begin{pmatrix}1\\0\end{pmatrix} + c_2(t)M\begin{pmatrix}0\\1\end{pmatrix}
\\&= c_1(t)2\begin{pmatrix}1\\0\end{pmatrix} + c_2(t)
\left(2\begin{pmatrix}0\\1\end{pmatrix} + \begin{pmatrix}1\\0\end{pmatrix}\right).
\end{align*}
Equating coefficients, we get:
\begin{align*}
c_1' = 2c_2 + c_2\\
c_2' = 2c_2.
\end{align*}
We we solve the second equation, we see $$c_2(t) = A_2e^{2t}$$. Plugging this into the first equation we get:
\begin{align*}
c_1' = 2c_2 + A_2e^{2t}.
\end{align*}
This can be solved (for example) using integrating factors:
\begin{align*}
c_1(t) = A_2te^{2t} + A_1e^{2t}.
\end{align*}
So then:
\begin{align*}
\boldsymbol{x}(t)
&= (A_2te^{2t} + A_1e^{2t})\begin{pmatrix}1\\0\end{pmatrix}
+ A_2e^{2t}\begin{pmatrix}0\\1\end{pmatrix}
\\[4mm]&=A_1e^{2t}\begin{pmatrix}1\\0\end{pmatrix}
+ A_2e^{2t}\left(t\begin{pmatrix}1\\0\end{pmatrix}
+ \begin{pmatrix}0\\1\end{pmatrix}\right).
\end{align*}

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L8V6

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

#### Video Notes

More generically, let's say we have the ODE:
\begin{align*}
\boldsymbol{x}'(t) = M\boldsymbol{x}(t)
\end{align*}
with one eigenvalue, $$\lambda$$, one lineraly independent eigenvector $$\boldsymbol{v}$$, and one generalized eigenvector, $$\boldsymbol{w}$$. And recall that $$(M-\lambda I)\boldsymbol{w} =\boldsymbol{v}$$, that is: $$M\boldsymbol{w} = \lambda \boldsymbol{w} + \boldsymbol{v}$$.

Since $$\boldsymbol{v}$$ and $$\boldsymbol{w}$$ are linearly independent, we can write the solution vector as a linear combination of them:
\begin{align*}
\boldsymbol{x}(t)
= c_1(t)\boldsymbol{v} + c_2(t)\boldsymbol{w}.
\end{align*}
Plugging this into the differential equation:
\begin{align*}
\boldsymbol{x}(t)
= c_1'(t)\boldsymbol{v} + c_2'(t)\boldsymbol{w}
= c_1(t)M\boldsymbol{v} + c_2(t)M\boldsymbol{w}
= c_1(t)\lambda\boldsymbol{v} + c_2(t)(\lambda \boldsymbol{w} + \boldsymbol{v}).
\end{align*}
Equating coefficients, this gives:
\begin{align*}
c_1' = \lambda c_1 + c_2\\
c_2' = \lambda c_2.
\end{align*}
Solving for $$c_2$$ gives $$c_2(t) = A_2e^{\lambda t}$$ and plugging this into the first equation and solving for $$c_1$$ gives:
\begin{align*}
c_1(t)
= A_2 te^{2t} + A_1e^{2t}.
\end{align*}
So then:
\begin{align*}
\boldsymbol{x}(t)
&= c_1(t)\boldsymbol{v} + c_2(t)\boldsymbol{w}
\\&= (A_2 te^{2t} + A_1e^{2t})\boldsymbol{v} + A_2e^{\lambda t}\boldsymbol{w}
\\&= A_1e^{2t}\boldsymbol{v} + A_2e^{2t}(t\boldsymbol{v} + \boldsymbol{w}).
\end{align*}

Exercises

1. Find a solution to the initial value problem:
\begin{align*}
\boldsymbol{x}'(t)
\begin{pmatrix}
-2 & 1\\
1 & -2
\end{pmatrix}
\boldsymbol{x}(t),\hspace{.5in}
\boldsymbol{x}(0)=\begin{pmatrix}1\\1\end{pmatrix}.
\end{align*}

We saw in a previous video that the eigenvalues and eigenvectors are:
\begin{align*}
\lambda_1 = -3, \hspace{.25in}\boldsymbol{v}_1=\begin{pmatrix}1\\-1\end{pmatrix}\\
\lambda_2 = -1, \hspace{.25in}\boldsymbol{v}_2=\begin{pmatrix}1\\1\end{pmatrix}.
\end{align*}
So the general solution is:
\begin{align*}
\begin{pmatrix}x_1(t)\\x_2(t)\end{pmatrix}
=A_1e^{-3t}\begin{pmatrix}1\\-1\end{pmatrix} +
A_2e^{-t}\begin{pmatrix}1\\1\end{pmatrix}.
\end{align*}
To find $$A_1$$ and $$A_2$$ we solve:
\begin{align*}
\begin{pmatrix}1\\1\end{pmatrix}
=\boldsymbol{x}(0)
=A_1\begin{pmatrix}1\\-1\end{pmatrix} +
A_2\begin{pmatrix}1\\1\end{pmatrix}.
\end{align*}
So then the second equation says $$A_2 = 1+ A_1$$ and plugging this into $$1 = A_1 + A_2 = 1 + 2A_1$$. So then $$A_1 = 0$$ and $$A_2 = 1$$. So the particular solution to this IVP is:
\begin{align*}
\begin{pmatrix}x_1(t)\\x_2(t)\end{pmatrix}
=e^{-t}\begin{pmatrix}1\\1\end{pmatrix}.
\end{align*}

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L8E1

2. Solve the initial value problem:
\begin{align*}
\boldsymbol{x}'(t)=
\begin{pmatrix}
3 & -2\\
4 & -1
\end{pmatrix}
\boldsymbol{x}(t),\hspace{.25in}
\boldsymbol{x}(0)=\begin{pmatrix}1\\1\end{pmatrix}.
\end{align*}

We already found the eigenvalues and eigenvectors:
\begin{align*}
\lambda_1 = 1+2i, \hspace{.25in}\boldsymbol{v}_1=\begin{pmatrix}1\\1-i\end{pmatrix}\\
\lambda_2 = 1-2i, \hspace{.25in}\boldsymbol{v}_2=\begin{pmatrix}1\\1+i\end{pmatrix}.
\end{align*}
We need to find the real and imaginary parts of $$e^{(1+2i)t}\begin{pmatrix}1\\1-i\end{pmatrix}$$:
\begin{align*}
e^{(1+2i)t}\begin{pmatrix}1\\1-i\end{pmatrix}
&= e^t\left(\cos 2t + i \sin 2t\right)\left(
\begin{pmatrix}1\\1\end{pmatrix} +
i\begin{pmatrix}0\\-1\end{pmatrix}
\right)
\\[4mm]&=e^t\begin{pmatrix}\cos 2t\\ \cos 2t +\sin 2t\end{pmatrix}
+ie^t\begin{pmatrix}\sin 2t\\ -\cos 2t + \sin 2t\end{pmatrix}.
\end{align*}
So the general solution is:
\begin{align*}
\begin{pmatrix}x_1(t)\\x_2(t)\end{pmatrix}
=A_1 e^t \begin{pmatrix}\cos 2t\\ \cos 2t +\sin 2t\end{pmatrix} +
A_2 e^t \begin{pmatrix}\sin 2t\\ -\cos 2t + \sin 2t\end{pmatrix}.
\end{align*}
To find $$A_1$$ and $$A_2$$, we use the initial condition:
\begin{align*}
\begin{pmatrix}1\\1\end{pmatrix}
=\boldsymbol{x}(0)
= A_1\begin{pmatrix}1\\1\end{pmatrix} +
A_2\begin{pmatrix}0\\-1\end{pmatrix}.
\end{align*}
Thus, $$A_1 = 1$$ and $$1 = A_1 - A_2 = 1-A_2$$ and so $$A_2 = 0$$. Thus the solution to the IVP is:
\begin{align*}
\begin{pmatrix}x_1(t)\\x_2(t)\end{pmatrix}
=e^t \begin{pmatrix}\cos 2t\\ \cos 2t +\sin 2t\end{pmatrix}.
\end{align*}

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L8E2

3. Find the general solution to
\begin{align*}
\boldsymbol{x}'(t)
=
\begin{pmatrix}
0 & 1\\
-4 & 4
\end{pmatrix}
\boldsymbol{x}(t).
\end{align*}

We have seen that $$\lambda = 2$$ is the only eigenvalue and the only linearly independent eigenvector is $$\boldsymbol{v}=\begin{pmatrix}1\\2\end{pmatrix}.$$ Also, $$\boldsymbol{w} = \begin{pmatrix}0\\1\end{pmatrix}$$ is a generalized eigenvector. So the general solution is:
\begin{align*}
\begin{pmatrix}x_1(t)\\x_2(t)\end{pmatrix}
= A_1e^{2t}\begin{pmatrix}1\\2\end{pmatrix}
+ A_2e^{2t}\left(t\begin{pmatrix}1\\2\end{pmatrix} +
\begin{pmatrix}0\\1\end{pmatrix}\right).
\end{align*}

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L8E3

4. Solve the initial value problem
\begin{align*}
\boldsymbol{x}'(t)=
\begin{pmatrix}
1 & 0 & 0\\
-3 & -2 &3\\
0 & 0 & 1
\end{pmatrix}
\boldsymbol{x}(t),
\hspace{.5in}
\boldsymbol{x}(0)=\begin{pmatrix}-1\\0\\1\end{pmatrix}.
\end{align*}

We saw in a previous video that the eigenvalues and eigenvectors are:
\begin{align*}
\lambda_1 = -2&, \hspace{.25in}\boldsymbol{v}_1=\begin{pmatrix}0\\1\\0\end{pmatrix}\\[4mm]
\lambda_{2,3} = 1&, \hspace{.25in}\boldsymbol{v}_2=\begin{pmatrix}1\\0\\1\end{pmatrix},
\begin{pmatrix}-1\\1\\0\end{pmatrix}.
\end{align*}
So the general solution is:
\begin{align*}
\begin{pmatrix}x_1(t)\\x_2(t)\\x_3(t)\end{pmatrix}
=A_1e^{-2t}\begin{pmatrix}0\\1\\0\end{pmatrix}
+A_2e^{t}\begin{pmatrix}1\\0\\1\end{pmatrix}
+A_3e^{t}\begin{pmatrix}-1\\1\\0\end{pmatrix}.
\end{align*}
To find $$A_1$$, $$A_2$$, and $$A_3$$ we use the initial condition:
\begin{align*}
\begin{pmatrix}-1\\0\\1\end{pmatrix}
&=A_1\begin{pmatrix}0\\1\\0\end{pmatrix}+
A_2\begin{pmatrix}1\\0\\1\end{pmatrix}+
A_3\begin{pmatrix}-1\\1\\0\end{pmatrix}
\\[4mm]&=\begin{pmatrix}
0 & 1 & -1\\
1 & 0 & 1\\
0 & 1 & 0
\end{pmatrix}
\begin{pmatrix}A_1\\A_2\\A_3\end{pmatrix}.
\end{align*}
So we put this in an augmented matrix and row–reduce:
\begin{align*}
\left(\begin{array}{ccc|c}
0 & 1 & -1 & -1\\
1 & 0 & 1 & 0\\
0 & 1 & 0 & 1
\end{array}\right)
\xrightarrow{R1 \leftrightarrow R2}
\left(\begin{array}{ccc|c}
1 & 0 & 1 & 0\\
0 & 1 & -1 & -1\\
0 & 1 & 0 & 1
\end{array}\right)\\[4mm]
\xrightarrow{R3 := R3 - R2}
\left(\begin{array}{ccc|c}
1 & 0 & 1 & 0\\
0 & 1 & -1 & -1\\
0 & 0 & 1 & 2
\end{array}\right)\\[4mm]
\xrightarrow[R1 := R1 - R3]{R2 := R2 + R3}
\left(\begin{array}{ccc|c}
1 & 0 & 0 & -2\\
0 & 1 & 0 & 1\\
0 & 0 & 1 & 2
\end{array}\right).
\end{align*}
So $$A_1 = -2$$, $$A_2 = 1$$ and $$A_3 = 2$$. So then:
\begin{align*}
\begin{pmatrix}x_1(t)\\x_2(t)\\x_3(t)\end{pmatrix}
=-2e^{-2t}\begin{pmatrix}0\\1\\0\end{pmatrix}
+e^{t}\begin{pmatrix}1\\0\\1\end{pmatrix}
+2e^{t}\begin{pmatrix}-1\\1\\0\end{pmatrix}.
\end{align*}

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Linear Algebra for Math 308: L8E4

5. Solve the initial value problem
\begin{align*}
\boldsymbol{x}'(t)
\begin{pmatrix}
0 & -1 & 1\\
-3 & 0 & 2\\
-5 & -3 & 5
\end{pmatrix}
\boldsymbol{x}(t),
\hspace{.25in}
\boldsymbol{x}(0) = \begin{pmatrix}0\\1\\0\end{pmatrix}.
\end{align*}

We saw in a previous video that the eigenvalues and eigenvectors and generalized eigenvectors are are:
\begin{align*}
\lambda_1 = 1&, \hspace{.25in}\boldsymbol{v}_1=\begin{pmatrix}1\\1\\2\end{pmatrix}\\
\lambda_{2,3} = 2&, \hspace{.25in}\boldsymbol{v}_2=\begin{pmatrix}0\\1\\1\end{pmatrix},
\boldsymbol{w}=\begin{pmatrix}1\\-2\\0\end{pmatrix}.
\end{align*}
Note: $$\boldsymbol{w}$$ is a generalized eigenvector and satisfies the equation $$(M - 2I)\boldsymbol{w} =\boldsymbol{v}_2$$. We write the solution vector as:
\begin{align*}
\begin{pmatrix}x_1(t)\\x_2(t)\\x_3(t)\end{pmatrix}
=c_1(t)\begin{pmatrix}1\\1\\2\end{pmatrix}+
c_2(t)\begin{pmatrix}0\\1\\1\end{pmatrix}+
c_3(t)\begin{pmatrix}1\\-2\\0\end{pmatrix}.
\end{align*}
Plugging this into the ODE we find:
\begin{align*}
c_1'(t)\begin{pmatrix}1\\1\\1\end{pmatrix}+
c_2'(t)\begin{pmatrix}0\\1\\1\end{pmatrix}+
c_3'(t)\begin{pmatrix}1\\-2\\0\end{pmatrix}
= c_1(t)\begin{pmatrix}1\\1\\2\end{pmatrix}+
2c_2(t)\begin{pmatrix}0\\1\\1\end{pmatrix}+
c_3(t)\left(\begin{pmatrix}0\\1\\1\end{pmatrix}+
2\begin{pmatrix}1\\-2\\0\end{pmatrix}\right).
\end{align*}
So then:
\begin{align*}
c_1' = c_1\\
c_2' = 2c_2 + c_3\\
c_3' = 2c_3.
\end{align*}
Therefore:
\begin{align*}
c_1(t) &= A_1e^{t}\\
c_3(t) &= A_3e^{2t}.
\end{align*}
Now we can solve $$c_2$$:
\begin{align*}
c_2' - 2c_2 = A_3e^{2t}.
\end{align*}
We can solve this using an integrating factor with $$\mu(t) = e^{-2t}$$. Then $$c_2(t) = A_3 te^{2t} + A_2e^{2t}$$. So then the general solution is:
\begin{align*}
\begin{pmatrix}x_1(t)\\x_2(t)\\x_3(t)\end{pmatrix}
&=A_1e^{t}\begin{pmatrix}1\\1\\2\end{pmatrix}+
(A_3 te^{2t} + A_2e^{2t})\begin{pmatrix}0\\1\\1\end{pmatrix}+
A_3e^{2t}\begin{pmatrix}1\\-2\\0\end{pmatrix}
\\[4mm]&=A_1e^{t}\begin{pmatrix}1\\1\\2\end{pmatrix}
+ A_2e^{2t}\begin{pmatrix}0\\1\\1\end{pmatrix}
+ A_3e^{2t}\left(t\begin{pmatrix}0\\1\\1\end{pmatrix} +
\begin{pmatrix}1\\-2\\0\end{pmatrix}\right).
\end{align*}
We find the coefficients with the initial condition:
\begin{align*}
\begin{pmatrix}0\\1\\0\end{pmatrix}
=\boldsymbol{x}(0)
=A_1\begin{pmatrix}1\\1\\2\end{pmatrix}
+ A_2\begin{pmatrix}0\\1\\1\end{pmatrix}
+ A_3 \begin{pmatrix}1\\-2\\0\end{pmatrix}.
\end{align*}
We can solve this to get $$A_1 = 1$$, $$A_2 = -2$$ and $$A_3 = -1$$. So the solution to the initial value problem is:
\begin{align*}
\begin{pmatrix}x_1(t)\\x_2(t)\\x_3(t)\end{pmatrix}
=e^{t}\begin{pmatrix}1\\1\\2\end{pmatrix}
+ -2e^{2t}\begin{pmatrix}0\\1\\1\end{pmatrix}
- e^{2t}\left(t\begin{pmatrix}0\\1\\1\end{pmatrix} +
\begin{pmatrix}1\\-2\\0\end{pmatrix}\right).
\end{align*}

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Linear Algebra for Math 308: L8E5

6. Solve the initial value problem
\begin{align*}
\boldsymbol{x}'(t)
\begin{pmatrix}
5 & -1\\
3 & 1
\end{pmatrix}
\boldsymbol{x}(t),
\hspace{.25in}
\boldsymbol{x}(0) = \begin{pmatrix}1\\0\end{pmatrix}.
\end{align*}

We already found the eigenvalues and eigenvectors:
\begin{align*}
\lambda_1 = 2, \hspace{.25in}\boldsymbol{v}_1=\begin{pmatrix}1\\3\end{pmatrix}\\
\lambda_2 = 4, \hspace{.25in}\boldsymbol{v}_2=\begin{pmatrix}1\\1\end{pmatrix}.
\end{align*}
So the general solution is:
\begin{align*}
\begin{pmatrix}x_1(t)\\x_2(t)\end{pmatrix}
=A_1 e^{2t}\begin{pmatrix}1\\3\end{pmatrix}+
A_2 e^{4t}\begin{pmatrix}1\\1\end{pmatrix}.
\end{align*}
To find the coefficients we use the initial condition:
\begin{align*}
\begin{pmatrix}1\\0\end{pmatrix}
= A_1\begin{pmatrix}1\\3\end{pmatrix}+
A_2\begin{pmatrix}1\\1\end{pmatrix}.
\end{align*}
So then $$A_1 = -\frac{1}{2}$$ and $$A_2 = \frac{3}{2}$$. So then
\begin{align*}
\begin{pmatrix}x_1(t)\\x_2(t)\end{pmatrix}
=-\frac{1}{2} e^{2t}\begin{pmatrix}1\\3\end{pmatrix}+
\frac{3}{2} e^{4t}\begin{pmatrix}1\\1\end{pmatrix}.
\end{align*}

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Linear Algebra for Math 308: L8E6