Elementary Differential Equations

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First Order Linear Systems of Equations

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Solve the following system:
$$ \begin{align} x' &= 3x,\qquad &x(0) = -6 \\ y' &= -4x + y,\qquad &y(0) = 1 \end{align} $$ 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) $$ \begin{align} sX + 6 &= 3X \\ sY - 1 &= -4X + Y \end{align} $$ 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.

$$ \begin{align} (s - 3)X &= -6 \\ 4X + (s - 1)Y&= 1 \end{align} $$

(*)

Now we multiply the first equation by -4 and the second equation by (s - 3) and add the two equations to get $$ (s^2 - 4s + 3)Y = s + 21$$ Next we divide through by $ s^2 - 4s + 3 $ to solve for $ Y.$ $$ Y = \frac{s + 21}{s^2 - 4s + 3} $$ We can immediately solve for $ X $ from the first equation in (*). $$ X = \frac-6{s - 3} $$ 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 $$ \begin{align} x(t) &= -6\exp(3t) \\ y(t) &= 12\exp(3t) - 11\exp(t) \\ \end{align} $$ You may reload this page to generate additional examples.

Kansas State University

Mathematics Department

First Order Linear Systems of Equations

Additional Examples