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 1


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 $\tau\in\{0,c\}.$


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