Elementary Differential Equations

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First Order Linear Systems of Equations

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Solve the following system:
$$ \begin{align} x' &= x - 13y,\qquad &x(0) = -3 \\ y' &= x + 5y,\qquad &y(0) = 1 \end{align} $$ In what follows, we will use capital letter notation for the Laplace transform, i.e. $ {\mathcal L}\{x(t)\} = X(s)$ and $ {\mathcal L}\{y(t)\} = Y(s).$

STEP 1: Take the Laplace tranform of both sides (of both equations) $$ \begin{align} sX + 3 &= X - 13Y \\ sY - 1 &= X + 5Y \end{align} $$ STEP 2: Solve for $ X $ and $ Y. $

We first put $ X $ and $ Y $ on the left sides and the constants on the right sides of the equations.

$$ \begin{align} (s - 1)X + 13Y &= -3 \\ -X + (s - 5)Y&= 1 \end{align} $$

(*)

Now we multiply the second equation by (s - 1) and add the two equations to get $$ (s^2 - 6s + 18)Y = s - 4$$ Next we divide through by $ s^2 - 6s + 18 $ to solve for $ Y.$ $$ Y = \frac{s - 4}{s^2 - 6s + 18} $$ Now we multiply the second equation in (*) by -13 and the first equation in (*) by (s - 5) and add the two equations to get $$ (s^2 - 6s + 18)X = -3s + 2 $$ Finally we divide through by $ s^2 - 6s + 18 $ to solve for $ X.$ $$ X = \frac{-3s + 2}{s^2 - 6s + 18} $$ STEP 3: Take the inverse Laplace transforms to find $ x(t) $ and $ y(t).$

We use partial fractions, just as we have for the past several sections to obtain the solution functions $$ \begin{align} x(t) &= -3\exp(3t)\cos(3t) - (7/3)\exp(3t)\sin(3t) \\ y(t) &= \exp(3t)\cos(3t) - (1/3)\exp(3t)\sin(3t) \\ \end{align} $$ You may reload this page to generate additional examples.

Kansas State University

Mathematics Department

First Order Linear Systems of Equations

Additional Examples