Elementary Differential Equations

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First Order Linear Systems of Equations

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Solve the following system:
$$ \begin{align} x' &= 7x - 13y,\qquad &x(0) = -2 \\ y' &= 2x - 3y,\qquad &y(0) = -4 \end{align} $$ 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) $$ \begin{align} sX + 2 &= 7X - 13Y \\ sY + 4 &= 2X - 3Y \end{align} $$ 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.

$$ \begin{align} (s - 7)X + 13Y &= -2 \\ -2X + (s + 3)Y&= -4 \end{align} $$

(*)

Now we multiply the first equation by 2 and the second equation by (s - 7) and add the two equations to get $$ (s^2 - 4s + 5)Y = -4s + 24$$ Next we divide through by $ s^2 - 4s + 5 $ to solve for $ Y.$ $$ Y = \frac{-4s + 24}{s^2 - 4s + 5} $$ Now we multiply the second equation in (*) by -13 and the first equation in (*) by (s + 3) and add the two equations to get $$ (s^2 - 4s + 5)X = -2s + 46 $$ Finally we divide through by $ s^2 - 4s + 5 $ to solve for $ X.$ $$ X = \frac{-2s + 46}{s^2 - 4s + 5} $$ 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 $$ \begin{align} x(t) &= -2\exp(2t)\cos(t) + 42\exp(2t)\sin(t) \\ y(t) &= -4\exp(2t)\cos(t) + 16\exp(2t)\sin(t) \\ \end{align} $$ You may reload this page to generate additional examples.

Kansas State University

Mathematics Department

First Order Linear Systems of Equations

Additional Examples