MHB Is This Calculation of the Laplace Transform Correct?

[Drain Brain]

please check my work here

$\mathscr{L}[2\sin(bt)\sinh(bt)]$

I know that $\sinh(bt) = \frac{e^{bt}-e^{-bt}}{2}$$\mathscr{L}[2\sin(bt)\left(\frac{e^{bt}-e^{-bt}}{2}\right)]$

$\mathscr{L}[e^{bt}\sin(bt)-e^{-bt}\sin(bt)]$$\mathscr{L}[e^{bt}\sin(bt)]-\mathscr{L}[e^{-bt}\sin(bt)]$$\frac{b}{s^2+b^2}|_{s-->s-b} - \frac{b}{s^2+b^2}|_{s-->s+b}$$\frac{b}{(s-b)^2+b^2}-\frac{b}{(s+b)^2+b^2}$$\frac{b}{(s^2-2bs+2b^2)}-\frac{b}{(s^2+2bs+2b^2)}$
$\frac{b(s^2+2bs+2b^2)-b(s^2-2bs+2b^2)}{(s^2-2bs+2b^2)(s^2+2bs+2b^2)}$

$\frac{4b^2s}{(s^2-2bs+2b^2)(s^2+2bs+2b^2)}$---->>>final answer please check this.

if my answer is correct how Am I going to get the inverse laplace of it?

regards and thanks!

 

[evinda]

please check my work here

$\mathscr{L}[2\sin(bt)\sinh(bt)]$

I know that $\sinh(bt) = \frac{e^{bt}-e^{-bt}}{2}$$\mathscr{L}[2\sin(bt)\left(\frac{e^{bt}-e^{-bt}}{2}\right)]$

$\mathscr{L}[e^{bt}\sin(bt)-e^{-bt}\sin(bt)]$$\mathscr{L}[e^{bt}\sin(bt)]-\mathscr{L}[e^{-bt}\sin(bt)]$$\frac{b}{s^2+b^2}|_{s-->s-b} - \frac{b}{s^2+b^2}|_{s-->s+b}$$\frac{b}{(s-b)^2+b^2}-\frac{b}{(s+b)^2+b^2}$$\frac{b}{(s^2-2bs+2b^2)}-\frac{b}{(s^2+2bs+2b^2)}$
$\frac{b(s^2+2bs+2b^2)-b(s^2-2bs+2b^2)}{(s^2-2bs+2b^2)(s^2+2bs+2b^2)}$

$\frac{4b^2s}{(s^2-2bs+2b^2)(s^2+2bs+2b^2)}$---->>>final answer please check this.

if my answer is correct how Am I going to get the inverse laplace of it?

regards and thanks!

Yes, it is right ! (Yes)

In order to get the inverse Laplace of it, use this:

$$F(s)=\frac{b}{(s-a)^2+b^2}$$

$$f(t)=\mathscr{L}^{-1} \{ F(s) \} =e^{at} \sin (bt)$$

• $F(s)=\frac{b}{(s-b)^2+b^2}$
  
  $$f(t)=\mathscr{L}^{-1} \{ F(s) \} =e^{bt} \sin (bt)$$
  
• $F(s)=\frac{b}{(s+b)^2+b^2}$
  
  $$f(t)=\mathscr{L}^{-1} \{ F(s) \}=e^{-bt} \sin (bt)$$

So, we have:

$$\mathscr{L}^{-1} \{ \frac{b}{(s-b)^2+b^2} - \frac{b}{(s+b)^2+b^2} \}=e^{bt} \sin (bt)-e^{-bt} \sin (bt)= \sin (bt) (e^{bt}-e^{-bt}) \\ =2 \sin (bt) \left ( \frac{e^{bt}-e^{-bt}}{2} \right )=2 \sin (bt) \sinh (bt)$$