non homogeneous linear differential equations using Laplace

Worked Example

Solve the following initial value problem:

y^{″} - 3 y^{'} + 2 y = e^{3 t}, y (0) = 1, y^{'} (0) = 0

L {y^{″}} - 3 L {y^{'}} + 2 L {y} = L {e^{3 t}}

Using the identity $L {e^{a t}} = \frac{1}{s - a}$ :

(s^{2} Y (s) - s y (0) - y^{'} (0)) - 3 (s Y (s) - y (0)) + 2 Y (s) = \frac{1}{s - 3}

(s^{2} Y (s) - s - 0) - 3 (s Y (s) - 1) + 2 Y (s) = \frac{1}{s - 3}

(s^{2} - 3 s + 2) Y (s) - s + 3 = \frac{1}{s - 3}

(s^{2} - 3 s + 2) Y (s) = \frac{1}{s - 3} + s - 3

(s - 1) (s - 2) Y (s) = \frac{1 + (s - 3)^{2}}{s - 3} = \frac{s^{2} - 6 s + 10}{s - 3}

Y (s) = \frac{s^{2} - 6 s + 10}{(s - 1) (s - 2) (s - 3)}

Set up the decomposition for distinct linear factors:

\frac{s^{2} - 6 s + 10}{(s - 1) (s - 2) (s - 3)} = \frac{A}{s - 1} + \frac{B}{s - 2} + \frac{C}{s - 3}

Solve for coefficients using the cover-up method:

$A = \frac{s^{2} - 6 s + 10}{(s - 2) (s - 3)} |_{s = 1} = \frac{1 - 6 + 10}{(- 1) (- 2)} = \frac{5}{2}$
$B = \frac{s^{2} - 6 s + 10}{(s - 1) (s - 3)} |_{s = 2} = \frac{4 - 12 + 10}{(1) (- 1)} = - 2$
$C = \frac{s^{2} - 6 s + 10}{(s - 1) (s - 2)} |_{s = 3} = \frac{9 - 18 + 10}{(2) (1)} = \frac{1}{2}$

Substitute $A, B,$ and $C$ back into $Y (s)$ :

Y (s) = \frac{5 / 2}{s - 1} - \frac{2}{s - 2} + \frac{1 / 2}{s - 3}

Apply $L^{- 1} {\frac{1}{s - a}} = e^{a t}$ term-by-term:

y (t) = \frac{5}{2} e^{t} - 2 e^{2 t} + \frac{1}{2} e^{3 t}