Kansas State University

Mathematics Department

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First Order Linear Systems of Equations

Additional Examples

Solve the following system:
$$ \begin{align} x' &= 3x + 20y,\qquad &x(0) = -1 \\ y' &= -2x + 7y,\qquad &y(0) = -1 \end{align} $$ 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) $$ \begin{align} sX + 1 &= 3X + 20Y \\ sY + 1 &= -2X + 7Y \end{align} $$ 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.

$$ \begin{align} (s - 3)X - 20Y &= -1 \\ 2X + (s - 7)Y&= -1 \end{align} $$ (*)
Now we multiply the first equation by -2 and the second equation by (s - 3) and add the two equations to get $$ (s^2 - 10s + 61)Y = -s + 5$$ Next we divide through by $ s^2 - 10s + 61 $ to solve for $ Y.$ $$ Y = \frac{-s + 5}{s^2 - 10s + 61} $$ Now we multiply the second equation in (*) by 20 and the first equation in (*) by (s - 7) and add the two equations to get $$ (s^2 - 10s + 61)X = -s - 13 $$ Finally we divide through by $ s^2 - 10s + 61 $ to solve for $ X.$ $$ X = \frac{-s - 13}{s^2 - 10s + 61} $$ 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 $$ \begin{align} x(t) &= -\exp(5t)\cos(6t) - 3\exp(5t)\sin(6t) \\ y(t) &= -\exp(5t)\cos(6t) \\ \end{align} $$ You may reload this page to generate additional examples.

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©2010, 2014 Andrew G. Bennett