Solve the following differential equation using Laplace Transforms:

$$y'' - y' - 6y = 0 $$

Where y is a function of t and has the following initial conditions:

$$y(0) = 1 $$

$$y'(0) = -1 $$

We start by taking the Laplace Transform of both sides of the equation:

$$\mathcal{L}(y'') - \mathcal{L}(y') - 6\mathcal{L}(y) = \mathcal{L}(0) $$

Using the definitions provided in our Laplace transform table we obtain the following:

$$[s^2 \mathcal{L}(y) - sy(0) - y'(0)] - [s\mathcal{L}(y) - y(0)] - [6\mathcal{L}(y)] = 0 $$

After collecting the Laplace transform terms and plugging in values for our given initial conditions, we get:

$$\mathcal{L}(y)[s^2 - s - 6] - s(1) - (-1) + 1 = 0 $$

$$\mathcal{L}(y)[s^2 - s - 6] - s + 2 = 0 $$

$$\mathcal{L}(y)[s^2 - s - 6] = s - 2 $$

$$\mathcal{L}(y) = \frac{s-2}{s^2 - s - 6} $$

In order to obtain an equation for y(t), we need to take the inverse Laplace transform of the right-hand side of the equation, such that:

$$y(t) = \mathcal{L^{-1}}[\frac{s-2}{s^2 - s - 6}] \qquad (Eqn \; 1)$$

In its current form, we can't easily determine the inverse Laplace transform of the term inside the brackets. We need to figure out a way to make it resemble something in the 2nd and 4th columns of our Laplace Transform table. Let's start by factoring the denominator:

$$y(t) = \mathcal{L^{-1}}[\frac{s-2}{(s-3)(s+2)}] $$

Now we will use partial fraction decomposition on the term inside the brackets:

Side work (Partial fraction decomposition)

$$\frac{s-2}{(s-3)(s+2)} = \frac{A}{s-3} + \frac{B}{s+2} $$ $$s-2 = A(s+2) + B(s-3) $$ letting s = -2: $$-4 = -5B $$ $$B = \frac{4}{5} $$ letting s = 3: $$1 = 5A $$ $$A = \frac{1}{5} $$ Therefore: $$\frac{s-2}{(s-3)(s+2)} = \frac{1}{5(s-3)} + \frac{4}{5(s+2)} $$

We substitute the above result of our partial fraction decomposition into equation #1 and get the following:

$$y(t) = \mathcal{L^{-1}}[\frac{1}{5(s-3)} + \frac{4}{5(s+2)}] \qquad (Eqn \; 2)$$

Go ahead and distribute the inverse Laplace symbol:

$$y(t) = \mathcal{L^{-1}}[\frac{1}{5(s-3)}] + \mathcal{L^{-1}}[\frac{4}{5(s+2)}] $$

We can also take the constant terms outside of the inverse Laplace terms:

$$y(t) = \frac{1}{5} \mathcal{L^{-1}}[\frac{1}{s-3}] + \frac{4}{5} \mathcal{L^{-1}}[\frac{1}{s+2}] $$

We now compare the terms inside of the brackets to our Laplace transform table and you may notice similarities with item number two (e^(at)). If we apply that relationship to the two inverse Laplace transform terms in our above expression, we get:

Final solution:

$$y(t) = \frac{1}{5} e^{3t} + \frac{4}{5} e^{-2t} $$

As you may have gathered by now, the one downside to solving differential equations using Laplace transforms is that it requires a high degree of intuition that only comes from doing a bunch of them and becoming familiar with the various relationships in the table (as well as drawing on numerous algebraic manipulation techniques). With some practice, you will start to get a "feel" for where the expressions are heading in terms of making them look like something in the table.