Kansas State University

Mathematics Department

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First Order Linear Systems of Equations

Additional Examples

Solve the following system:
$$ \begin{align} x' &= 7x + 17y,\qquad &x(0) = 2 \\ y' &= -x - y,\qquad &y(0) = -4 \end{align} $$ 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) $$ \begin{align} sX - 2 &= 7X + 17Y \\ sY + 4 &= -X - Y \end{align} $$ 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.

$$ \begin{align} (s - 7)X - 17Y &= 2 \\ X + (s + 1)Y&= -4 \end{align} $$ (*)
Now we multiply the first equation by -1 and the second equation by (s - 7) and add the two equations to get $$ (s^2 - 6s + 10)Y = -4s + 26$$ Next we divide through by $ s^2 - 6s + 10 $ to solve for $ Y.$ $$ Y = \frac{-4s + 26}{s^2 - 6s + 10} $$ Now we multiply the second equation in (*) by 17 and the first equation in (*) by (s + 1) and add the two equations to get $$ (s^2 - 6s + 10)X = 2s - 66 $$ Finally we divide through by $ s^2 - 6s + 10 $ to solve for $ X.$ $$ X = \frac{2s - 66}{s^2 - 6s + 10} $$ 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 $$ \begin{align} x(t) &= 2\exp(3t)\cos(t) - 60\exp(3t)\sin(t) \\ y(t) &= -4\exp(3t)\cos(t) + 14\exp(3t)\sin(t) \\ \end{align} $$ You may reload this page to generate additional examples.

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©2010, 2014 Andrew G. Bennett