How to solve initial value problems with Laplace transforms

Laplace transforms turn the derivatives in an initial value problem into expressions involving one unknown, $Y (s)$ . We can solve for that unknown with algebra, then take the inverse transform to find $y (t)$ . Let’s work through the steps on a second-order equation.

Transform the derivatives and use the initial values

Find $y (t)$ for the following equation and initial conditions.

\begin{array}{l} y'' - 10 y' + 9 y = 5 t \\ y (0) = - 1, y' (0) = 2 \end{array}

Write $Y (s) = ℒ (y (t))$ . The derivative transforms include the initial values, so we can substitute them as we transform each term.

\begin{array}{l} ℒ (y') = s Y - y (0) \\ ℒ (y'') = s^{2} Y - s y (0) - y' (0) \\ ℒ (5 t) = \frac{5}{s^{2}} \end{array}

Here the first derivative transforms to $s Y + 1$ and the second transforms to $s^{2} Y + s - 2$ . Bring the transformed terms together.

\begin{array}{l} (s^{2} Y + s - 2) - 10 (s Y + 1) + 9 Y \\ = \frac{5}{s^{2}} \end{array}

Solve for the transformed function

Collect the terms containing the unknown transform, then factor the quadratic in the denominator.

\begin{array}{l} (s^{2} - 10 s + 9) Y = \frac{5}{s^{2}} + 12 - s \\ Y (s) = \frac{5 + 12 s^{2} - s^{3}}{s^{2} (s - 9) (s - 1)} \end{array}

Decompose into partial fractions

The factor $s$ occurs twice, so we need terms with denominators $s$ and $s^{2}$ . The other factors each contribute one term.

\begin{array}{l} Y (s) = \frac{A}{s} + \frac{B}{s^{2}} \\ + \frac{C}{s - 9} + \frac{D}{s - 1} \end{array}

Multiply by $s^{2} (s - 9) (s - 1)$ . Setting $s = 0$ isolates $B$ . Setting $s = 9$ isolates $C$ , and setting $s = 1$ isolates $D$ .

\begin{array}{l} B = \frac{5}{(- 9) (- 1)} = \frac{5}{9} \\ C = \frac{248}{9^{2} (9 - 1)} = \frac{31}{81} \\ D = \frac{16}{1^{2} (1 - 9)} = - 2 \end{array}

The coefficient of $s^{3}$ in the original numerator is $- 1$ . In the combined fractions, the $A$ , $C$ , and $D$ terms each contribute to that coefficient.

\begin{array}{l} A + C + D = - 1 \\ A = - 1 - \frac{31}{81} + 2 = \frac{50}{81} \end{array}

Take the inverse Laplace transform

The inverse transforms of $1 / s$ , $1 / s^{2}$ , and $1 / (s - a)$ are $1$ , $t$ , and $e^{a t}$ , respectively. Apply those pairs to the four terms.

\begin{array}{l} y (t) = \frac{50}{81} + \frac{5}{9} t \\ + \frac{31}{81} e^{9 t} - 2 e^{t} \end{array}

Check the initial values and the equation

\begin{array}{l} y (0) = \frac{50 + 31}{81} - 2 = - 1 \\ y' (0) = \frac{5}{9} + \frac{31}{9} - 2 = 2 \end{array}

The exponentials satisfy the homogeneous equation because $9$ and $1$ are roots of $r^{2} - 10 r + 9$ . Applying $y'' - 10 y' + 9 y$ to the remaining terms $50 / 81 + 5 t / 9$ gives $- 50 / 9 + 50 / 9 + 5 t = 5 t$ . This checks the forcing term.

Shift the time variable for initial values at another point

For initial values at $t = t_{0}$ , set $τ = t - t_{0}$ and write $u (τ) = y (τ + t_{0})$ . The given values now specify $u (0)$ and $u' (0)$ . Transform the equation in $τ$ , then substitute $τ = t - t_{0}$ into the solution.

Learn mathKrista KingMay 12, 2019math, learn online, online course, online math, differential equations, laplace transforms, table of laplace transforms, definition of the laplace transform, initial value problems, IVPs, initial conditions