# Inverse Laplace Transform

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#### cbluvaluva

##### Newbie level 1
i am having trouble with some aspects of Laplace transform, could anyone help me with this problem

Find the inverse Laplace transform.
4(s+1)/s²-16

1/(s²+9)

1/(s+1)(s+2)
Laplace is new to me so your help would be greatly appreciated.

#### claudiocamera

##### Full Member level 4
4(s+1)/s²-16 here you have poles in s=+4 and s=-4 so you aplly fraction expansion of the form A/(s+4) + B/(s-4) find A anb and take the inverse of A/(s+4) and B/(s-4) which is A exp(-4t)u(t) + Bexp(4t)u(t).

1/(s²+9) the inverse Laplace transform of this function can be found directly from tables: It is sin(3t)u(t).

1/(s+1)(s+2) Here again you should use fraction expansion : A/(s+1) + B/(s+2) find A and B and take the inverse and you will find Aexp(-t)u(t) + Bexp(-2t)u(t).

You shoul take a look at fraction expansion method. I hope have helped.

#### dreamcard

##### Member level 2
Intuition or experience is ok,better way is to use the definition of this kind of transformations,in the textbook

##### Junior Member level 2
any rational function can be expanded and after that simple inverses exist.

#### hamidr_karami

##### Banned
hi

ilaplace(4*(s+1)/(s*s-16)) ---------> (5/2 exp(4 t) + 3/2 exp(-4 t))*u(t)

ilaplace(1/(s*s+9)) ---------> 1/9*(9^(1/2))*sin(9^(1/2)*t) *u(t)

ilaplace(1/(s+1)(s+2)) ----------> exp(-t)-exp(-2*t)

bye

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