MHB How to Solve Laplace Transforms with a Fractional Term?

[rannasquaer]

How to solve the transforms below

\[ \mathscr{L}^{-1} \frac{a(s+2 \lambda)+b}{(s+ \lambda)^2- \omega^2} \]

 

[I like Serena]

How to solve the transforms below

\[ \mathscr{L}^{-1} \frac{a(s+2 \lambda)+b}{(s+ \lambda)^2- \omega^2} \]

The table of Laplace transforms lists that $\mathscr{L}^{-1} \frac{s+\alpha}{(s+\alpha)^2+\omega^2} = e^{-\alpha t}\cos(\omega t)\cdot u(t)$ and $\mathscr{L}^{-1} \frac{\omega}{(s+\alpha)^2+\omega^2} = e^{-\alpha t}\sin(\omega t)\cdot u(t)$.

Can we use those to find the requested transform?

 

[rannasquaer]

The table of Laplace transforms lists that $\mathscr{L}^{-1} \frac{s+\alpha}{(s+\alpha)^2+\omega^2} = e^{-\alpha t}\cos(\omega t)\cdot u(t)$ and $\mathscr{L}^{-1} \frac{\omega}{(s+\alpha)^2+\omega^2} = e^{-\alpha t}\sin(\omega t)\cdot u(t)$.

Can we use those to find the requested transform?

I think yes, if I rewrite like

\[ \mathscr{L}^{-1} \frac{a(s+\lambda)}{(s+ \lambda)^2- \omega^2} + \mathscr{L}^{-1} \frac{b + \lambda a}{(s+ \lambda)^2- \omega^2} \]

but I have \[ (s+\lambda)^2-\omega^2 \] and not \[ (s+\lambda)^2+\omega^2 \]

The table of Laplace transforms lists that \[ \mathscr{L}^{-1} \frac{\alpha}{s^2- \alpha^2} = \sin h(\alpha t).u(t) \] and \[ \mathscr{L}^{-1} \frac{s}{s^2- \alpha^2} = \cos h(\alpha t).u(t) \]

I do not know what to do now

 

[I like Serena]

I think yes, if I rewrite like

\[ \mathscr{L}^{-1} \frac{a(s+\lambda)}{(s+ \lambda)^2- \omega^2} + \mathscr{L}^{-1} \frac{b + \lambda a}{(s+ \lambda)^2- \omega^2} \]

but I have \[ (s+\lambda)^2-\omega^2 \] and not \[ (s+\lambda)^2+\omega^2 \]

The table of Laplace transforms lists that \[ \mathscr{L}^{-1} \frac{\alpha}{s^2- \alpha^2} = \sin h(\alpha t).u(t) \] and \[ \mathscr{L}^{-1} \frac{s}{s^2- \alpha^2} = \cos h(\alpha t).u(t) \]

I do not know what to do now

Right. I meant the $\cosh$ and $\sinh$ versions.

Also note that $\mathscr{L}^{-1} F(s-\alpha)=e^{\alpha t}f(t)$.

So we can do:
\[ \mathscr{L}^{-1} \frac{a(s+\lambda)}{(s+ \lambda)^2- \omega^2} + \mathscr{L}^{-1} \frac{b + \lambda a}{(s+ \lambda)^2- \omega^2} =a e^{-\lambda t}\sinh(\omega t) \cdot u(t)+ \frac{b+\lambda a}{\omega}e^{-\lambda t}\cosh(\omega t) \cdot u(t)\]
And if we want to, we can rewrite it using $\sinh x= \frac 12(e^x-e^{-x})$ and $\cosh x=\frac 12(e^x + e^{-x})$.

 

[rannasquaer]

Great, I understood how to continue to do the math. Thank you!