Laplace transform of a function

[Drain Brain]

please help me solve this problem

$\mathscr{L}\{e^{3a-2bt}\}$

here's my attempt

$\mathscr{L}\{e^{3a}\cdot e^{-2bt}\}$ from here I couldn't continue

I looked up my table of transform but nothing matches the problem above. I'm not sure if the first shift formula would work here. please help.

regards :)

- - - Updated - - -

Oh my! I didn't notice that $e^a$ is just a constant. I can pull it out and take the laplace transform of $e^{-2bt}$

 

[Prove It]

please help me solve this problem

$\mathscr{L}\{e^{3a-2bt}\}$

here's my attempt

$\mathscr{L}\{e^{3a}\cdot e^{-2bt}\}$ from here I couldn't continue

I looked up my table of transform but nothing matches the problem above. I'm not sure if the first shift formula would work here. please help.

regards :)

- - - Updated - - -

Oh my! I didn't notice that $e^a$ is just a constant. I can pull it out and take the laplace transform of $e^{-2bt}$

Yes you pull out $\displaystyle \begin{align*} \mathrm{e}^{3a} \end{align*}$ as a constant factor. The remaining function should be easy to find the Laplace Transform of :)

 

[Drain Brain]

yes solved it already.

I have another problem here

$\mathscr{L}\{(t^2-3)^2\}$

what's the first step here. I couldn't see any transform that matches the function from my table.

regards.

 

[Prove It]

yes solved it already.

I have another problem here

$\mathscr{L}\{(t^2-3)^2\}$

what's the first step here. I couldn't see any transform that matches the function from my table.

regards.

Expand out the brackets.

 

[Drain Brain]

Expand out the brackets.

$\mathscr{L}\{t^4-6t^2+9\}$

now $\frac{4!}{s^{4+1}}-\frac{6(2!)}{s^{2+1}}+\frac{9}{s}$

$\frac{24}{s^{5}}-\frac{12}{s^3}+\frac{9}{s}$ is this correct?

 

[Prove It]

$\mathscr{L}\{t^4-6t^2+9\}$

now $\frac{4!}{s^{4+1}}-\frac{6(2!)}{s^{2+1}}+\frac{9}{s}$

$\frac{24}{s^{5}}-\frac{12}{s^3}+\frac{9}{s}$ is this correct?

Looks good to me :)