# initial problem ODE solve with Laplace Transformation

Consider the initial value problem

$\displaystyle{ x''(t) +3x'(t)+2x(t)= t + (ae^{-(t-a)}-t) H(t-a)=:f(t) \quad ; x(0)=x'(0)=0}$ where $a>0$ is constant and $H$ is the Heaviside function.

Applying the Laplace transformation with $\mathcal{L \{x(t) \}}=X(s)$ and $\mathcal{L \{f(t) \} } =F(s)$ and using the initial conditions and after some calculation we get

$\displaystyle{ X(s) =\frac{1}{(s+1)(s+2)} F(s) }$ from which we get $\displaystyle{ x(t) = \mathcal{L^{-1} \{\frac{1}{(s+1)(s+2)} {F(s)}\}} }$

and so it is $\displaystyle{ x(t) = \mathcal{ L^{-1} \{\frac{F(s)}{s+1} \} } -\mathcal{ L^{-1} \{\frac{F(s)}{s+2} \} } = e^{-t}*f(t) - e^{-2t}*f(t)= e^{-t}(1-e^{-t})*f(t) }$

where $*$ is the operator of convolution.

How can I continue now to do as least calculations as it is possible?

edit: I am interested also if there is a simpler way (using Laplace Transformation) tell me.

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Cannot you explicitly calculate $F(s)$, and use its expression for the anti-transformation, instead of using convolution? – enzotib Oct 20 '12 at 8:40

I think you're better off just evaluating the inverse directly. That is, evaluate $F(s)$ and divide by $(s+1)(s+2)$, and then use a table or evaluate the Bromwich contour in the complex $s$ plane to get $x(t)$.

You compute $F(s)$ from

\begin{align}F(s) &= \int_0^{\infty} dt \: [t + (a e^{t-a}-t) H(t-a)] e^{-s t}\\ &= \frac{1-e^{-a s}}{s^2} + a \frac{e^{-a s}}{s+1} - a \frac{e^{-a s}}{s} \end{align}

(Please check my work, although I do believe I am correct.)

The LT of the solution is then

$$X(s) = \frac{F(s)}{(s+1)(s+2)} = \frac{1-e^{-a s}}{s^2 (s+1)(s+2)} + a \frac{e^{-a s}}{(s+1)^2 (s+2)} - a \frac{e^{-a s}}{s(s+1)(s+2)}$$

The ILT of this is straightforward, if a little messy.

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Thank you very much for your answer! – passenger Apr 27 '13 at 16:35

The best way is to compute the convolution integral in the s-plane.Not in the time plane. Use the identity: $$\mathcal{L\{f(t)\ast g(t)\}}=\mathcal{L \{\int_o^tf(t-\tau)g(\tau)d\tau \} } =F(s)G(s)$$, where F(s) and G(s) are the Laplace transforms of f(t) and g(t) respectively.

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