Question

Could you please explain how to solve differential equations using matrices. I understand that it somehow involves eigenvalues and eigenvectors. I am not familiar with either eigenvalues or eigenvectors....

Answer 1

Maybe you want to understand how to solve systems of differential equations using matrices.

Let we have the system of differential equations $Y' = AY$. These steps help you to solve this system:

1. Find the matrix $P$ that diagonalises $A$, i.e. the matrix $P$ such that $D = P^{-1}AP$, where $D$ is a diagonal matrix.

2. Make the change of variable $Y = PU, Y' = PU'$. We then have

$$Y' = AY \Leftrightarrow PU' = APU \Leftrightarrow U' = P^{-1}APU \Leftrightarrow U' = DU$$

3. Solve the system $U' = DU$ for $U$.

4. The solution to the system of differential equation is $Y = PU$, where $U$ is the solution to $U' = DU$.

Let's see how it works. First we have to find the matrix $P$, i.e. we have to find the eigenvalues and eigenvectors of the matrix $A$. Let us have the matrix

$$A = \left( \begin{array}{cc} 1 & 1 \\ 4 & -2 \end{array} \right)$$

Then

$$|A - \lambda I| = \left| \begin{array}{cc} 1 - \lambda & 1 \\ 4 & -2 - \lambda \end{array} \right| = \lambda^2 + \lambda - 6 = (\lambda - 2)(\lambda + 3) = 0$$

so that $A$ has two eigenvalues: $\lambda = 2$ and $\lambda = -3$.

Let $X = \binom{X}{y}$

$$AX = 2X \Leftrightarrow \left\{ \begin{array}{l} -x + y = 0 \\ 4x - 4y = 0 \end{array} \right. \Leftrightarrow x = y \Leftrightarrow X = \alpha \binom{1}{1}$$ $$AX = -3X \Leftrightarrow \left\{ \begin{array}{l} 4x + y = 0 \\ 4x - 4y = 0 \end{array} \right. \Leftrightarrow y = -4x \Leftrightarrow X = \beta \binom{1}{-4}$$

The matrix $P$ is therefore $P = \left( \begin{array}{cc} 1 & 1 \\ 1 & -4 \end{array} \right)$ and the diagonal matrix $D$ is $D = \left( \begin{array}{cc} 2 & 0 \\ 0 & -3 \end{array} \right)$.

Let $U = \binom{u_1(x)}{u_2(x)}$. The matrix equality $U' = DU$ is equivalent to

$$\left\{ \begin{array}{l} u_1'(x) = 2u_1(x) \\ u_2'(x) = -3u_2(x)' \end{array} \right.$$

which is easily solved:

$$\left\{ \begin{array}{l} u_1'(x) = 2u_1(x) \\ u_2'(x) = -3u_2(x) \end{array} \right. \Leftrightarrow u_1(x) = Ae^{2x}, u_2(x) = Be^{-3x} \Leftrightarrow U = \binom{Ae^{2x}}{Be^{-3x}}$$

$Y = PU$ is the solution to $Y' = AY$ and we have

$$Y = PU = \left( \begin{array}{cc} 1 & 1 \\ 1 & -4 \end{array} \right) \binom{Ae^{2x}}{Be^{-3x}} = \binom{Ae^{2x} + Be^{-3x}}{Ae^{2x} - 4Be^{-3x}}$$

If you are also given initial conditions, i.e. conditions of the form

$$Y(0) = \left( \begin{array}{c} u(0) \\ v(0) \end{array} \right) = \left( \begin{array}{c} 1 \\ -1 \end{array} \right)$$

you can find the value of $A$ and the value of $B$ :

$$Y(0) = \left( \begin{array}{c} 1 \\ -1 \end{array} \right) \Leftrightarrow \left\{ \begin{array}{l} A + B = 1 \\ A - 4B = -1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} A = \frac{3}{5} \\ B = \frac{2}{5} \end{array} \right.$$

and finally the solution to the initial value problem is $Y = \frac{1}{5} \left( \frac{3e^{2x} + 2e^{-3x}}{3e^{2x} - 8e^{-3x}} \right)$