Find the inverse Laplace transform?

[Success]

Homework Statement

Find the inverse Laplace transform of F(s)=(2s-3)/(s^2-4).

Homework Equations

I don't want to find the answer by looking at the Table.
F(s)=2s/(s^2-4)-3/(s^2-4)

The Attempt at a Solution

The answer is f(t)=2 cosh 2t - (3/2) sinh 2t.

 

[Ray Vickson]

Homework Statement

Find the inverse Laplace transform of F(s)=(2s-3)/(s^2-4).

Homework Equations

I don't want to find the answer by looking at the Table.
F(s)=2s/(s^2-4)-3/(s^2-4)

The Attempt at a Solution

The answer is f(t)=2 cosh 2t - (3/2) sinh 2t.

Show your work. You are required to make an effort before receiving help here.

 

[Mandelbroth]

Homework Statement

Find the inverse Laplace transform of F(s)=(2s-3)/(s^2-4).

Homework Equations

I don't want to find the answer by looking at the Table.
F(s)=2s/(s^2-4)-3/(s^2-4)

The Attempt at a Solution

The answer is f(t)=2 cosh 2t - (3/2) sinh 2t.

Show some work. Then we can help.

Edit: I wasn't fast enough.

 

[Success]

F(s)=2s/(s^2-4)-3/(s^2-4)
A/(s+2)+B/(s-2)-A/(s+2)+B/(s-2)

 

[HallsofIvy]

Alright, and now what? You said you wanted to find the inverse Laplace Transform without "looking at the table". There is a very complex formula for finding an inverse Laplace Transform directly but that is exceptionally difficult to use! That is why there are tables!

So I guess you could do what you have already tried to do, write the transform as a sum of "partial fractions" and try to remember (without looking at the table!) what the original function for each is.

You have, however, done the 'partial fractions' incorrectly. From what you have you should have seen that your "A/(s+2)" and "-A/(s+ 2)" will cancel while "B/(s-2)" and "+B/(s- 2)" will add. You want [itex](2s- 3)/(s^2- 4)= (2s- 3)/((s+2)(s-2))= A/(s+2)+ B/(s-2).

 

[Mandelbroth]

Alright, and now what? You said you wanted to find the inverse Laplace Transform without "looking at the table". There is a very complex formula for finding an inverse Laplace Transform directly but that is exceptionally difficult to use! That is why there are tables!

I see what you did there. Very complex formula.

For the sake of making what we mean clear to the OP, your inverse Laplace transform will be given by $$\mathcal{L}^{-1}\left\{\frac{2s-3}{s^2-4}\right\}(t)=\frac{1}{2\pi i}\lim_{b\rightarrow\infty}\int_{a-bi}^{a+bi}e^{st}\left(\frac{2s-3}{s^2-4}\right)~ds,$$ for $a$ greater than 2. This is typically considered undesirable to have to compute.

 

[Success]

Mandel, is that the only way to solve this problem?

 

[Success]

Also, what to do after you get A/(s+2)+B/(s-2)?

 

[vela]

Find A and B.

 

[Success]

Never mind. I got it using the table. Thanks for the help, everyone.