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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?

Thanks in advance!

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?2012-10-20

2 Answers 2