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I'm currently solving a differential equation via Laplace transforms and am unsure on how to apply the reverse transform to one of the terms. I have $L[F(x)]=f(s)$ and currently trying to perform:

$$L^{-1}\left[\frac{B}{A-s^2}\right]$$

I've tried to take the negative of this and try and match it with one in a table, but have been unsuccessful. I know the answer is

$$\frac{B e^{-\sqrt{A}}(e^{2\sqrt{A}}-A)}{2\sqrt{A}}$$

but cannot get the workings correctly (determined the answer from mathematica, but cant get the steps right).

Can anyone help with which transform to use in the table?

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  • $\begingroup$ Do you have any conditions on $A?$ $\endgroup$ – Adrian Keister Jul 27 '18 at 16:15
  • $\begingroup$ No conditions, A is just strictly some arbitrary constant. $\endgroup$ – MathHelper123 Jul 27 '18 at 16:16
  • $\begingroup$ Note Isham found a sign error in my previous answer, which I've now corrected. $\endgroup$ – Adrian Keister Jul 27 '18 at 16:47
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We have three possibilities, depending on the sign of $A$. First, let's assume $A>0.$ Then we have \begin{align*} \mathcal{L}^{-1}\left(\frac{B}{A-s^2}\right)&=B\,\mathcal{L}^{-1}\left(\frac{1}{\left(\sqrt{A}-s\right)\left(\sqrt{A}+s\right)}\right) \\ &=B\,\mathcal{L}^{-1}\left(\frac{1}{2\sqrt{A}}\left(\frac{1}{\sqrt{A}+s}+\frac{1}{\sqrt{A}-s}\right)\right) \\ &=\frac{B}{2\sqrt{A}}\,\mathcal{L}^{-1}\left(\frac{1}{\sqrt{A}+s}+\frac{1}{\sqrt{A}-s}\right)\\ &=\frac{B}{2\sqrt{A}}\left(e^{-\sqrt{A}\,t}-e^{\sqrt{A}\,t}\right)\\ &=-\frac{B}{\sqrt{A}}\,\sinh\left(\sqrt{A}\,t\right). \end{align*} You probably have unit step functions in there, technically, though you might not need them.

Now let's suppose $A<0.$ Let $C^2=|A|.$ So $C=\sqrt{|A|}.$ Then we have \begin{align*} \mathcal{L}^{-1}\left(\frac{B}{A-s^2}\right)&=\mathcal{L}^{-1}\left(\frac{B}{-|A|-s^2}\right) \\ &=-\frac{B}{C}\,\mathcal{L}^{-1}\left(\frac{C}{C^2+s^2}\right) \\ &=-\frac{B}{C}\,\sin(C\,t) \\ &=-\frac{B}{\sqrt{|A|}}\,\sin\left(\sqrt{|A|}\,t\right). \end{align*}

Finally, suppose $A=0.$ Then you have merely

\begin{align*} \mathcal{L}^{-1}\left(\frac{B}{A-s^2}\right)&=-B\,\mathcal{L}^{-1}\left(\frac{1}{s^2}\right) \\ &=-B\,t. \end{align*}

So, your final answer can be written this way: $$\mathcal{L}^{-1}\left(\frac{B}{A-s^2}\right)=\begin{cases}-\dfrac{B}{\sqrt{A}}\,\sinh\left(\sqrt{A}\,t\right),\quad &A>0\\ -B\, t,\quad &A=0 \\-\dfrac{B}{\sqrt{-A}}\,\sin\left(\sqrt{-A}\,t\right),\quad &A<0 \end{cases}. $$

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  • $\begingroup$ Thank you so much, this helps immensely! $\endgroup$ – MathHelper123 Jul 27 '18 at 16:34
  • $\begingroup$ Adrian you have a sign mistake ...it should be Sinh not cosh $\endgroup$ – Isham Jul 27 '18 at 16:37
  • $\begingroup$ @Isham: Mathematica disagrees with you, I think. I could well have made a sign error, but it's not apparent. $\endgroup$ – Adrian Keister Jul 27 '18 at 16:39
  • $\begingroup$ @Isham: Hmm. Looks like you're right. Mistake has to be in the partial fraction decomposition. Hang on.. $\endgroup$ – Adrian Keister Jul 27 '18 at 16:44
  • $\begingroup$ @Isham: Found it, thanks! $\endgroup$ – Adrian Keister Jul 27 '18 at 16:47
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$$f(t)=\mathcal{L^{-1}}(B/(A-s^2))$$ $$f(t)=\frac {B}{2\sqrt A}\mathcal{L^{-1}}(\frac 1{(\sqrt A-s)}+\frac 1{(\sqrt A+s)})$$ $$f(t)=\frac {B}{2\sqrt A}\mathcal{L^{-1}}(-\frac 1{(s-\sqrt A)}+\frac 1{(s+\sqrt A)})$$ Look at the table of lapalce transform $$f(t)=\frac {B}{2\sqrt A}(-e^{\sqrt At}+e^{-\sqrt At})$$ $$f(t)=-\frac {B}{\sqrt A}\sinh(\sqrt At)$$

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  • $\begingroup$ For $A,B\ne 0$ .....otherwise for $A=0 \implies f(t)=-Bt$ $\endgroup$ – Isham Jul 27 '18 at 17:45

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