When you take the transform, you are no longer solving the differential equation for x(t), you are now solving a differential equation for the Laplace transform. After you solve that equation, you need to reverse the transform to find x(t)[/tex], which is the original problem.<br />
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The Laplace transform is given by the formula<br />
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\tilde{f}(s)=\int_{0}^{\infty}e^{-st}f(t)dt<br />
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for example, the laplace transform for the derivative is given by<br />
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\widetilde{f&#039;}(s)=\int_{0}^{\infty}e^{-st}f&#039;(t)dt=-sf(0)+s\int_{0}^{\infty}e^{-st}f(t)dt<br />
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so<br />
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\widetilde{f&#039;}(s)=-sf(0)+s\tilde{f}(s)<br />
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using this result<br />
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\widetilde{f&#039;&#039;}(s)=-sf&#039;(0)+s\widetilde{f&#039;}(s)=-sf&#039;(0)-s^{2}f(0)+s^{2}\tilde{f}(s)<br />
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now, you only need to transform the whole equation, solve for \tilde{x}(s) and find the inverse transform as Galileo said (using tables). For further reference check Boyce DiPrima ode's book.<br />
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Ah... and your equation represents an harmonic oscilator under the influence of an external force f(t). The I must be a constant. The amplitude of a pulse.<br />
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Lets do 1. for example.<br />
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given x(0)=0,x&#039;(0)=0 implies that<br />
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\widetilde{x&#039;}(s)=s\tilde{x}(s),\widetilde{x&#039;&#039;}(t)=s^{2}\tilde{x}(s)<br />
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and <br />
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\mathcal{L}[\delta(t-c)]=\int_{0}^{\infty}e^{-st}\delta(t-c)dt<br />
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wich by definition of the delta function f(y)=\int f(x)\delta(x-y)dx yields<br />
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\mathcal{L}[\delta(t-c)]=e^{-sc}<br />
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so \mathcal{L}[\delta(t)]=1. Now, transforming the whole equation, <br />
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\widetilde{mx&#039;&#039;(t)+kx(t)}=\widetilde{I\delta}(t)<br />
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due linearity<br />
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\widetilde{mx&#039;&#039;(t)+kx(t)}=m\widetilde{x&#039;&#039;}(t)+k\tilde{x}(t)=I\widetilde{\delta}(t)<br />
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substituting<br />
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(ms^{2}+k)\tilde{x}(s)=I<br />
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so<br />
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\tilde{x}(s)=\frac{I}{m}\frac{1}{s^{2}+\omega^{2}}<br />
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and given the fact that (try calculating this one, good exercise)<br />
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\mathcal{L}[sen(\omega t)]=\frac{\omega}{s^{2}+\omega^{2}}<br />
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implies<br />
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\tilde{x}(s)=\mathcal{L}[x(t)]=\frac{I}{m\omega}\mathcal{L}[sen(\omega t)]<br />
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Inverting<br />
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x(t)=\frac{I}{m\omega}sen(\omega t)<br />
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There is some details missing, but I am too tired and i think this show the main point. You need to use your initial conditions and the homogeneous solution. Anyone here is willing to complete the exercise?
Example problems attached.
