How to take Laplace transform $L\left\{\frac{\cos at - \cos bt}{b^2-a^2}\right\}$

Question

I need to verify that

$$L\left\{\frac{\cos at - \cos bt}{b^2-a^2}\right\} = \frac{s}{(s^2+a^2)(s^2+b^2)}$$

for $a^2\neq b^2$

I found one solution that just says that:

$$L\left\{\frac{\cos at - \cos bt}{b^2-a^2}\right\} = \frac{\frac{s}{s^2+a^2}-\frac{s}{s^2+b^2}}{b^2-a^2} = \frac{s}{(s^2+a^2)(s^2+b^2)}$$

I understood that it's because $L\{\cos at\} = \frac{s}{s^2+a^2}$ and $L\{-\cos bt\} = -\frac{s}{s^2+b^2}$ and I can separate into sums because the Laplace transformation is linear. But what about $$\frac{1}{b^2-a^2}$$ ?

Shouldn't I apply the transformation to that too? Even if it does not depend on $t$, it should have a transformation, because even $L\{1\}$ is not trivial. I think I should've done:

$$L\left\{\frac{\cos at - \cos bt}{b^2-a^2}\right\} = (L\{\cos at\} - L\{\cos bt\})L\left\{\frac{1}{b^2-a^2}\right\}$$

Answer 1

The fact that $\mathcal{L}$ is a linear operator also implies that for any $\lambda\in\mathbb{R}$ $$ \mathcal{L}(\lambda g) = \lambda\cdot \mathcal{L}(g) $$ and you just have to consider $\lambda=\frac{1}{b^2-a^2}$.