Kansas State University

Mathematics Department

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First Order Linear Systems of Equations

Additional Examples

Solve the following system:
$$ \begin{align} x' &= 2x - 20y,\qquad &x(0) = 5 \\ y' &= 2x + 6y,\qquad &y(0) = -3 \end{align} $$ 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) $$ \begin{align} sX - 5 &= 2X - 20Y \\ sY + 3 &= 2X + 6Y \end{align} $$ 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.

$$ \begin{align} (s - 2)X + 20Y &= 5 \\ -2X + (s - 6)Y&= -3 \end{align} $$ (*)
Now we multiply the first equation by 2 and the second equation by (s - 2) and add the two equations to get $$ (s^2 - 8s + 52)Y = -3s + 16$$ Next we divide through by $ s^2 - 8s + 52 $ to solve for $ Y.$ $$ Y = \frac{-3s + 16}{s^2 - 8s + 52} $$ Now we multiply the second equation in (*) by -20 and the first equation in (*) by (s - 6) and add the two equations to get $$ (s^2 - 8s + 52)X = 5s + 30 $$ Finally we divide through by $ s^2 - 8s + 52 $ to solve for $ X.$ $$ X = \frac{5s + 30}{s^2 - 8s + 52} $$ 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 $$ \begin{align} x(t) &= 5\exp(4t)\cos(6t) + (25/3)\exp(4t)\sin(6t) \\ y(t) &= -3\exp(4t)\cos(6t) + (2/3)\exp(4t)\sin(6t) \\ \end{align} $$ You may reload this page to generate additional examples.

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©2010, 2014 Andrew G. Bennett