Solving systems of differential equations

2 Solving systems of differential equations

The Laplace transform method is also well suited to solving systems of differential equations. A simple example will illustrate the technique.

Let $x (t), y (t)$ be two independent functions which satisfy the coupled differential equations

\begin{array}{rcl} \frac{d x}{d t} + y & = & e^{- t} \\ \frac{d y}{d t} - x & = & 3 e^{- t} \\ x (0) & = & 0, y (0) = 1 \end{array}

Now, using a traditional approach, we could try to eliminate one of the unknown functions from this system: for example, from the first:

$\frac{d y}{d t} = - e^{- t} - \frac{d^{2} x}{d t^{2}} (taking the derivative and rearranging)$

This can then be substituted in the second equation: $\frac{d y}{d t} - x = 3 e^{- t}$ , to give:

$- \frac{d^{2} x}{d t^{2}} - x = 4 e^{- t}$

which can then be solved in the normal way (either using the complementary function/particular integral approach or else the Laplace transform approach.) However, this approach is not workable if we have large numbers of first order differential equations to deal with. Let us instead use the Laplace transform directly.

If we use the notation that

$L {x (t)} = X (s) and L {y (t)} = Y (s)$

then, by taking the Laplace transform of every term in the given differential equations, we obtain:

\begin{array}{rcl} - x (0) + s X (s) + Y (s) & = & \frac{1}{s + 1} \\ - y (0) + s Y (s) - X (s) & = & \frac{3}{s + 1} \end{array}

which, using the initial conditions and rearranging gives

\begin{array}{rcl} s X (s) + Y (s) & = & \frac{1}{s + 1} \\ - X (s) + s Y (s) & = & \frac{s + 4}{s + 1} \end{array}

Key Point 13

Taking the Laplace transform converts a system of differential equations

into a system of algebraic simultaneous equations.

We can solve these algebraic equations (in $X (s)$ and $Y (s)$ ) using a variety of techniques (inverse matrix; Cramer’s determinant method etc.) Here we will use Cramer’s method.

\begin{array}{rcl} X (s) = \frac{|\begin{matrix} \frac{1}{s + 1} & 1 \\ \frac{s + 4}{s + 1} & s \end{matrix}|}{|\begin{matrix} s & 1 \\ - 1 & s \end{matrix}|} & = & \frac{\frac{s}{s + 1} - \frac{s + 4}{s + 1}}{s^{2} + 1} \\ = & \frac{- 4}{(s^{2} + 1) (s + 1)} = \frac{2 (s - 1)}{s^{2} + 1} - \frac{2}{s + 1} \end{array}

and

\begin{array}{rcl} Y (s) = \frac{|\begin{matrix} s & \frac{1}{s + 1} \\ - 1 & \frac{s + 4}{s + 1} \end{matrix}|}{|\begin{matrix} s & 1 \\ - 1 & s \end{matrix}|} & = & \frac{\frac{s (s + 4)}{s + 1} + \frac{1}{s + 1}}{s^{2} + 1} \\ = & \frac{s^{2} + 4 s + 1}{(s^{2} + 1) (s + 1)} = - \frac{1}{s + 1} + \frac{2 (s + 1)}{s^{2} + 1} \end{array}

The last lines in each case having been obtained using partial fractions. We can now invert $X (s), Y (s)$ to find $x (t), y (t)$ :

\begin{array}{rcl} x (t) = L^{- 1} {X (s)} & = & 2 L^{- 1} {\frac{s}{s^{2} + 1}} - 2 L^{- 1} {\frac{1}{s^{2} + 1}} - 2 L^{- 1} {\frac{1}{s + 1}} \\ = & (2 cos t - 2 sin t - 2 e^{- t}) u (t) \\ y (t) = L^{- 1} {Y (s)} & = & - L^{- 1} {\frac{1}{s + 1}} + 2 L^{- 1} {\frac{s}{s^{2} + 1}} + 2 L^{- 1} {\frac{1}{s^{2} + 1}} \\ = & (- e^{- t} + 2 cos t + 2 sin t) u (t) \end{array}

(Note that once the solution for $x (t)$ is found the solution for $y (t)$ may be easier to obtain by substituting in the differential equation: $y = e^{- t} - \frac{d x}{d t}$ rather than using Laplace transforms.)

Task!

Use the Laplace transform to solve the coupled differential equations:

$\frac{d y}{d t} - x = 0, \frac{d x}{d t} + y = 1, x (0) = - 1, y (0) = 1$

Begin by obtaining a system of algebraic equations for $X (s)$ and $Y (s)$ :

Writing $L {x (t)} = X (s)$ and $L {y (t)} = Y (s)$ you should obtain the set of transformed equations

$- 1 + s Y (s) - X (s) = 0$

$1 + s X (s) + Y (s) = \frac{1}{s}$

which, when re-arranged, are

$- X (s) + s Y (s) = 1$

$s X (s) + Y (s) = \frac{1 - s}{s}$

Now solve these equations for $X (s)$ and $Y (s)$ :

$X (s) = - \frac{s}{1 + s^{2}} Y (s) = \frac{1}{s} - \frac{1}{1 + s^{2}}$

Now find the required solution by obtaining the inverse Laplace transforms:

You should obtain $x (t) = - cos t . u (t)$ and $y (t) = (1 - sin t) . u (t)$ . This follows since

$L^{- 1} {- \frac{s}{1 + s^{2}}} = - cos t . u (t) L^{- 1} {\frac{1}{s}} = u (t) L^{- 1} {- \frac{1}{1 + s^{2}}} = - sin t . u (t)$

Exercises

Solve the given system of differential equations for the initial conditions specified.
1. $\frac{d x}{d t} = y \frac{d y}{d t} = x x (0) = 1 y (0) = 0$
2. $\frac{d x}{d t} = 4 x - 2 y \frac{d y}{d t} = 5 x + 2 y x (0) = 2 y (0) = - 2$
The Laplace transform can also be used to solve a pair of coupled second order differential equations.
Solve, for the given initial conditions,

$\frac{d^{2} x}{d t^{2}} = y + sin t x (0) = 1 x^{'} (0) = 0$

$\frac{d^{2} y}{d t^{2}} = - \frac{d x}{d t} + cos t y (0) = - 1 y^{'} (0) = - 1$

(Note that the initial conditions on each of $x (t)$ and $y (t)$ are needed in the second order situation.)

1. $x = cosh t, y = sinh t$
2. $x = e^{3 t} (2 cos 3 t + 2 sin 3 t), y = e^{3 t} (- 2 cos 3 t + 4 sin 3 t)$
$x = cos t, y = - cos t - sin t$

2 Solving systems of differential equations

Key Point 13

Task!

Answer

Answer

Answer

Exercises

Answer