Kansas State University

Mathematics Department

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First Order Linear Systems of Equations

Additional Examples

Solve the following system:
$$ \begin{align} x' &= 6x + 5y,\qquad &x(0) = 1 \\ y' &= -5x,\qquad &y(0) = -6 \end{align} $$ 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) $$ \begin{align} sX - 1 &= 6X + 5Y \\ sY + 6 &= -5X \end{align} $$ 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.

$$ \begin{align} (s - 6)X - 5Y &= 1 \\ 5X + sY&= -6 \end{align} $$ (*)
Now we multiply the first equation by -5 and the second equation by (s - 6) and add the two equations to get $$ (s^2 - 6s + 25)Y = -6s + 31$$ Next we divide through by $ s^2 - 6s + 25 $ to solve for $ Y.$ $$ Y = \frac{-6s + 31}{s^2 - 6s + 25} $$ Now we multiply the second equation in (*) by 5 and the first equation in (*) by s and add the two equations to get $$ (s^2 - 6s + 25)X = s - 30 $$ Finally we divide through by $ s^2 - 6s + 25 $ to solve for $ X.$ $$ X = \frac{s - 30}{s^2 - 6s + 25} $$ 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 $$ \begin{align} x(t) &= \exp(3t)\cos(4t) - (27/4)\exp(3t)\sin(4t) \\ y(t) &= -6\exp(3t)\cos(4t) + (13/4)\exp(3t)\sin(4t) \\ \end{align} $$ You may reload this page to generate additional examples.

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©2010, 2014 Andrew G. Bennett