System of Differential Equation | Differential Equations

System of Differential Equation

Sharan Sajiv Menon - November 16th, 2021

This article shows how to solve a first-order linear homogenous system of differential equation with constant coefficents.

Example 1: Solve the following Differential Equation with the Initial conditions

y'=\frac{1}{6} \begin{pmatrix} 7 & 2 \\ -2 & 2\end{pmatrix}y, \quad y(0)=\begin{pmatrix} 0\\-3\end{pmatrix}

Solution:

The general solution is of the following form.

y=c_1e^{\lambda_1{t}}E_1 + c_2e^{\lambda_2{t}}E_2 + ... c_ne^{\lambda_nt}E_n

Where $\lambda$ is the eigenvalues and $E$ is the eigenvector

Start by simplifying the equation

y'=\begin{pmatrix} 7/6 & 1/3 \\ -1/3 & 1/3\end{pmatrix}y, \quad y(0)=\begin{pmatrix} 0\\-3\end{pmatrix}

To solve this equation, we need to find the eigenvalues and eigenvectors of the matrix. Start by finding eigenvalues first.

Programming languages like MATLAB and Python can find the eigenvectors/eigenvalues automatically. A calculator like a TI-84 can do many operations to simplify the calculations. Use a Linear Algebra system to get the eigenvectors.

\begin{pmatrix} 7/6 & 1/3 \\ -1/3 & 1/3\end{pmatrix} - \lambda\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix} (7-6\lambda)/6 & 1/3 \\ -1/3 & (1-3\lambda)/3\end{pmatrix}

Now. take the determinant of that matrix.

\begin{vmatrix} \frac{7-6\lambda}{6} & 1/3 \\ -1/3 & \frac{1-3\lambda}{3} \end{vmatrix} = \frac{7-27\lambda-18\lambda^2}{18} + \frac{1}{9}=18\lambda^2-27\lambda+9=0

Solve the equation for $\lambda$ to get $\lambda=1, \frac{1}{2}$ . Now, finding the eigenvectors involve solving the following equation

(A-\lambda_i)x=0

Where $A$ is our initial matrix and $\lambda_i$ is our eigenvalue. There are 2 eigenvalues, therefore we have 2 eigenvectors.

Start by finding $(A-\lambda_i)$

\begin{pmatrix} 7/6 & 1/3 \\ -1/3 & 1/3\end{pmatrix} - 1\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}=\begin{pmatrix} 1/6 & 1/3 \\ -1/3 & -2/3\end{pmatrix}

Now, solve the following equation.

\begin{pmatrix} 1/6 & 1/3 \\ -1/3 & -2/3\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}

Above is a system of equations. It can be solved by taking the reduced row echelon form of $AB$ , where $B$ is the zeros-matrix. Solving this first system gets us the following eigenvector

\begin{pmatrix}-2\\1\end{pmatrix}, \quad \lambda=1

Following the same process for $\lambda=0.5$ , (make sure to replace the 1 with 0.5), we get the following eigenvector

\begin{pmatrix}-1\\2\end{pmatrix}, \quad \lambda=\frac{1}{2}

Noting the general form of the solution mentioned at the beginning, our general solution is the following

y= c_1e^t\begin{pmatrix}-2\\1\end{pmatrix} + c_2e^{0.5t}\begin{pmatrix}-1\\2\end{pmatrix}

To solve the IVP, just plug in the initial conditions into the problem.

\begin{pmatrix}0\\-3\end{pmatrix}= c_1e^0\begin{pmatrix}-2\\1\end{pmatrix} + c_2e^{0.5*0}\begin{pmatrix}-1\\2\end{pmatrix} \to \begin{pmatrix}0\\-3\end{pmatrix}= \begin{pmatrix} -2c_1-c_2 \\ c_1+c_2 \end{pmatrix}

We have another linear system of equations, which can again be solved by taking the inverse-matrix or the reduced-row echelon form. Solving this system and we get $c_1=1$ and $c_2=-1$ .

Therefore, our final solution is the following.

y= e^t\begin{pmatrix}-2\\1\end{pmatrix} -2e^{0.5t}\begin{pmatrix}-1\\2\end{pmatrix}