Ask your own question, for FREE!
Mathematics 20 Online
OpenStudy (anonymous):

Find the inverse Laplace transform of e^(-5s)/(s^2+3s+2).

OpenStudy (anonymous):

this looks like a heaviside function

OpenStudy (anonymous):

How do I do it?

OpenStudy (anonymous):

separate the laplace and you have e^-(5s)*(1)/(s^2+3s+2)

OpenStudy (anonymous):

f(t) will be the inverse laplace transform of (1)/(s^2+3s+2)

OpenStudy (anonymous):

do you know the heaviside formula?

OpenStudy (anonymous):

No.

OpenStudy (anonymous):

ok the eqaution is L{f(t-a)U(t-a)}=e^(-as)F(s)

OpenStudy (anonymous):

i mean equation* now you see the equation, we will work backwards to find the LHS of the equation

OpenStudy (anonymous):

our U(t-a) will be U(t-5) because and you can see that from the formula e^(-5s) where a=-5

OpenStudy (anonymous):

now we have to work on finding f(t) from (1)/(s^2+3s+2)

OpenStudy (anonymous):

you have the table in front of you

OpenStudy (anonymous):

Okay, I get it.

OpenStudy (anonymous):

@Idealist, Heaviside function is the same as unit step function.

OpenStudy (anonymous):

How would you do this problem?

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!