Differential Equations and Linear Algebra, 4th Edition

10.4 Derivatives, Integrals, and Products of Transforms

The Laplace transform of the (initially unknown) solution of a differential equation is sometimes recognizable as the product of the transforms of two known functions. For example, when we transform the initial value problem

$$x^{″}+x=\cos t;x(0)=x^{\text{′}}(0)=0,$$

we get

$$X(s)=\frac{s}{{(s^2+1)}^2}=\frac{s}{s^2+1}\cdot \frac{1}{s^2+1}=\mathcal{L}\{\cos t\}\cdot \mathcal{L}\{\sin t\}.$$

This strongly suggests that there ought to be a way of combining the two functions $\sin t$ and $\cos t$ to obtain a function x(t) whose transform is the product of their transforms. But obviously x(t) is not simply the product of $\cos t$ and $\sin t$, because

$$\mathcal{L}\{\cos t\sin t\}=\mathcal{L}\left\{\frac{1}{2}\sin 2t\right\}=\frac{1}{s^2+4}\ne \frac{s}{{(s^2+1)}^2}.$$

Thus $\mathcal{L}\{\cos t\sin t\}\ne \mathcal{L}\{\cos t\}·\mathcal{L}\{\sin t\}$.

Theorem 1 of this section will tell us that the function

$$h(t)={\displaystyle {\int }_0^{}}$$