Systems of ODEs

Let $A$ be a $2 \times 2$ matrix with real and distinct eigenvalues. Consider the problem of solving a system of ODE's in matrix-vector form,

$$ \dfrac{d \textbf{x}(t)}{dt}= A \textbf{x}(t) $$

subject to a given initial condition $ \textbf{x}(0) = \begin{bmatrix} x_1(0) \\ x_2(0) \end{bmatrix} $.

You can solve it according to the following steps

First find the eigenvalues $\lambda_1,\lambda_2$ of $A$

For each eigenvalue $\lambda_1,\lambda_2$, find the associated eigenvector $\textbf{v},\textbf{u}$.

Find the constants $c_1,c_2$ so that $c_1 \textbf{v} + c_2 \textbf{u} = \textbf{x}(0) $.

The final solution has the form, $ \textbf{x}(t) = c_1 e^{ \lambda_1 t} \textbf{v} + c_2 e^{ \lambda_2 t} \textbf{u} $

Solve Initial Value Problem for System Of Differential Equations

Solve the following System Of Differential Equations, subject to the initial value

For all generated problems, the matrix $A$ will have two distinct real eigenvalues.

The solution is,

First recall that any solution to a system of ODEs of the form, $ \dfrac{d \textbf{x}(t)}{dt} = A \textbf{x}(t) $ (where $A$ is a $2 \times 2$ matrix with distinct eigenvalues) will have the form,

$$ \textbf{x}(t) = c_1 e^{ \lambda_1 t} \textbf{v} + c_2 e^{ \lambda_2 t} \textbf{u} $$

for some constants $c_1,c_2$ where $\lambda_1,\lambda_2$ are eigenvalues with corresponding eigenvectors $\textbf{v},\textbf{u}$.

We start by finding the eigenvalues/eigenvectors and then later find the specific constants using the given initial value, $\textbf{x}(0) = \begin{bmatrix} x_1(0)\\x_2(0) \end{bmatrix} $

We start by finding the eigenvalues which we do by finding roots of $ p(\lambda ) = \det ( A - \lambda I ) =0 $.

For us this gives, $$ $ \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} $

This gives us the eigenvalues, and .

Now we can find the eigenvector associated to this eigenvalue.

Lets start with

Set $ \textbf{v} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} $ and compute,

The two equations are redundant. We can see they are solved for $v_1 = $ and $v_2 = $ which gives .

Now we move onto the second eigenvalue,

Set $ \textbf{u} = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} $ and compute,

Again, the two equations are redundant. We can see they are solved for $u_1 = $ and $u_2 = $ which gives .

To summarize, we have found the following eigenvalue/eigenvector pairs;

Eigenvalue with eigenvector

This means that the solution has the form,

Now it remains to find the constants $c_1$ and $c_2$. To do this we need to solve,

You can solve this system to find,

Which gives you the final solution,

Solving Systems of ODEs Refresher

Let \(A\) be a \(2 \times 2\) matrix with real and distinct eigenvalues. Consider the problem of solving a system of ODE's in matrix-vector form,

$$ \dfrac{d \textbf{x}(t)}{dt}= A \textbf{x}(t) $$

subject to a given initial condition \( \textbf{x}(0) = \begin{bmatrix} x_1(0) \\ x_2(0) \end{bmatrix} \).

You can solve it according to the following steps

Solve Initial Value Problem for System Of Differential Equations

Solve the following System Of Differential Equations, subject to the initial value

The solution is,

First recall that any solution to a system of ODEs of the form, \( \dfrac{d \textbf{x}(t)}{dt} = A \textbf{x}(t) \) (where \(A\) is a \(2 \times 2\) matrix with distinct eigenvalues) will have the form,

$$ \textbf{x}(t) = c_1 e^{ \lambda_1 t} \textbf{v} + c_2 e^{ \lambda_2 t} \textbf{u} $$

for some constants \(c_1,c_2\) where \(\lambda_1,\lambda_2\) are eigenvalues with corresponding eigenvectors \(\textbf{v},\textbf{u}\).

We start by finding the eigenvalues/eigenvectors and then later find the specific constants using the given initial value, \(\textbf{x}(0) = \begin{bmatrix} x_1(0)\\x_2(0) \end{bmatrix} \)

We start by finding the eigenvalues which we do by finding roots of \( p(\lambda ) = \det ( A - \lambda I ) =0 \).

For us this gives, $$ \( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \)

This gives us the eigenvalues, and .

Now we can find the eigenvector associated to this eigenvalue.

Lets start with

Set \( \textbf{v} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} \) and compute,

The two equations are redundant. We can see they are solved for \(v_1 = \) and \(v_2 = \) which gives .

Now we move onto the second eigenvalue,

Set \( \textbf{u} = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} \) and compute,

Again, the two equations are redundant. We can see they are solved for \(u_1 = \) and \(u_2 = \) which gives .

To summarize, we have found the following eigenvalue/eigenvector pairs;

Eigenvalue with eigenvector

Eigenvalue with eigenvector

This means that the solution has the form,

Now it remains to find the constants \(c_1\) and \(c_2\). To do this we need to solve,

You can solve this system to find,

Which gives you the final solution,