Solve DE by Laplace transformation: $y'' + y = 8e^{-2t} \sin t$?

Question

The differential equation is as follows

$$\begin{cases} y'' + y = 8e^{-2t} \sin t \\ y(0)=0 \\y'(0)=0 \end{cases}$$

How do I solve it by Laplace transformation?

In my solution I've taken the Laplace transform of both sides and got to the solution of the transformed equation $$\mathcal L\{y\} = \frac{8}{[(s-2)^2+1](s^2+1)}$$ but I'm stuck, I don't know how to take the inverse Laplace transform to get the solution to the DE.

Answer 1

Hint

Apply Laplace transform to both sides of the equation to get $$ s^2 \tilde y -s \underbrace{y(0^+) - y'(0^+)}_{\text{Zero from the}\\\text{initial conditions}} + \tilde y = \frac{8}{(s+2)^2+1} \\[20 pt] s^2\tilde y + \tilde y = \frac{8}{(s+2)^2+1}$$ Where $\tilde y = \mathcal L\{y\}$. Now you can easily solve for $\tilde y$ and do an inverse Laplace transform to find the solution $y$.

By simplifying the solution you'll get $$\tilde y = \frac{8}{[(s+2)^2+1](s^2+1)}$$ and by doing partial fraction decomposition $$\tilde y = \frac{As+B}{(s+2)^2+1}+\frac{Cs+D}{s^2+1} = \frac{(As+B)[(s+2)^2+1]+(Cs+D)(s^2+1)}{[(s+2)^2+1](s^2+1)}$$ Now just find the values of $A,B,C,D$ and then take the inverse Laplace of the decomposed function.

Answer 2

There are two methods that can be used. The first being the convolution and the second being the longer, more standard, version. Let the Laplace transform be given by $$\mathcal{L}\{y(t)\} = f(s) = \int_{0}^{\infty} e^{-s t} \, y(t) \, dt.$$ Since, $$\mathcal{L}\{ y'' + y\} = s^2 \, f - s y(0) - y'(0) + f = (s^2 + 1) \, f,$$ then by convolution the solution takes the form: \begin{align} y(t) = 8 \, \int_{0}^{t} \cos(t-u) \, e^{-2 u} \, \sin(u) \, du. \end{align} This is seen by \begin{align} (s^2 + 1) \, f &= \mathcal{L}\{8 \, e^{-2t} \, \sin(t) \} \\ f &= \frac{1}{s^2 + 1} \, \mathcal{L}\{8 \, e^{-2t} \, \sin(t) \} \\ &= \mathcal{L}\{\cos(t)\} \cdot \mathcal{L}\{8 \, e^{-2t} \, \sin(t) \} \end{align} and after the inversion, by convolution, the integral form, which can be calculated, is obtained.

The more standard version is: \begin{align} (s^2 + 1) \, f &= \mathcal{L}\{8 \, e^{-2t} \, \sin(t) \} = \frac{8}{(s+2)^2 + 1} \\ f &= \frac{8}{(s^2 + 1) \, ((s+2)^2 + 1)} \\ &= \frac{1-s}{s^2 + 1} + \frac{s+3}{(s+2)^2 + 1} \\ &= \frac{1}{s^2 + 1} - \frac{s}{s^2+1} + \frac{s+2}{(s+2)^2 + 1} + \frac{1}{(s+2)^2 + 1} \end{align} and leads to $$y(t) = \sin t - \cos t + e^{-2 t} \, (\sin t + \cos t).$$

Both methods yield the same result after some calculations are applied for the convolution case.