# Second order linear ODE equal to pulse funcion with Laplace transform method

The question is to solve the following initial conditions problem with the Laplace transform method.

$$f'' + 2f' -3f = \begin{cases} 1, \ 0 \leq t < c \\ 0, \ t \geq c \end{cases}; f(0) = f'(0) = 0$$

What I did was apply the Laplace transform to both sides so we get:

$$\mathcal{L}\{ f \}(s) \cdot (s^2 + 2s - 3) = \int_0^c e^{-st} dt = \frac{1-e^{-sc}}{s} \implies \mathcal{L}\{ f \} = \frac{1 - e^{-sc}}{s(s+3)(s-1)}$$

I have no idea how to continue from here. I wasn't able to figure out the inverse transform from the basic properties (Linearity, derivative, primitive, frequency translation, time translation, etc).

Thanks in advance for any responses.

## 1 Answer

You should decompose the rational function in partial fractions $$\frac{1}{s(s+3)(s-1)}=-\frac{1}{3s}+\frac{1}{12(s+3)}+\frac{1}{4(s-1)}$$ then consider the inverse Laplace transform $$\mathcal{L}^{-1}\left\{\frac{e^{-s\tau}}{s+s_0}\right\}=e^{-s_0(t-\tau)}H(t-\tau)$$ where $$H$$ is the unit step function.
Apply this formula for $$s_0\in\{0,3,-1\}$$ and $$\tau\in\{0,c\}.$$