# Find the Inverse Laplace Transform?

Find the inverse Laplace transform of:

(2s^2 -2s - 3) / (s + 1)(s^2 -2s +4)

Given:

?1{(2s2 ?2s ? 3) / [(s + 1)(s2 ? 2s + 4)]}

You can split this up into partial fractions:

(2s2 ? 2s ? 3) / [(s + 1)(s2 ? 2s + 4)] = A/(s + 1) + (Bs + C)/(s2 ?2s + 4)

? 2s2 ? 2s ? 3 = A(s2 ? 2s + 4) + (Bs + C)(s + 1)

? 2s2 ? 2s ? 3 = A(s2 ? 2s + 4) + Bs(s + 1) + C(s + 1)

If s = -1: 2 + 2 ?3 = 7A + 0 + 0

? 1 = 7A

? A = 1/7

If s = 0: -3 = 4A + C

? -3 = 4(1/7) + C

? C = -25/7

If s = 1 (you could really choose any number here): 2 ? 2 ?3 = 3A + 2B + 2C

? -3 = 3(1/7) + 2B + 2(-25/7)

? 13/7

That means the above is equivalent to:

?1{(1/7)(1/(s +?1)) + ((13s ? 25)/7)/(s2 ? 2s + 4)}

Complete the square to rewrite s2 ?2s + 4:

?1{(1/7)(1/(s + 1)) + ((13s ? 25)/7)/((s ?1)2?+ (√3)2)}

The Laplace transform is a linear operator, so we can rewrite that as:

(1/7)?1{1/(s + 1)} + (1/7)?1{(13s ? 25)/((s ? 1)2 + (√3)2)}

= (1/7)[?1{1/(s + 1)} + ?1{(13s ? 25)/((s ? 1)2 + (√3)2)}]

We can now split this into two problems.

For ?1{1/(s + 1)}, we (should) know that the inverse Laplace transform 1/(s ?a) = e^(at). Thus, the inverse Laplace transform of 1/(s + 1) is e^-t.

For ?1{(13s ? 25)/((s ? 1)2 + (√3)2)}, we'll have to change some things. First, you should notice that this is similar in resemblance to ?{e^(at)cos(bt)} = (s ?a)/[(s ?a)2?+ b2] and ?{e^(at)sin(bt)} = b/[(s ? a)2 + b2]. Therefore, we'll have to rewrite the numerator a bit.

We want to rewrite the numerator so that it contains s ?1. How do we do that? Just replace s ?1 where s is and compensate:

13(s ?1) ? 25 = 13s ?13 ? 25 = 13s ?38

Since this needs to be 13s ?25, we'll have to add 13 to it to compensate.

13(s ? 1) ? 25 + 13 = 13(s ?1) ? 12

You should verify that this new numerator is equal to the old one.

We can now rewrite that inverse Laplace as:

?1{(13(s ? 1) ? 12)/((s ? 1)2 + (√3)2)}

Which is (due to the Laplace transform being linear):

13?1{(s ? 1)/((s ? 1)2 + (√3)2)} ?1{12/((s ? 1)2 + (√3)2)}

Now we just need to rewrite 12 with √3 in it.

12 = 12(√3/√3)

Rationalizing the denominator, we get: 12√3√3 / 3 = 4√3√3.

Thus, ?1{12/((s ? 1)2 + (√3)2)} = ?1{4√(3)√(3)/((s ? 1)2 + (√3)2)}

The inverse Laplace transform of 13?1{(s ? 1)/((s ? 1)2 + (√3)2)} is 13e^(t)cos(√(3)t).

The inverse Laplace transform of 4√(3)?1{√(3)/((s ? 1)2 + (√3)2)} is 4√(3)e^(t)sin(√(3)t).

Putting it all together:

?1{(2s2 ? 2s ? 3) / [(s + 1)(s2 ? 2s + 4)]}

= (1/7)[?1{1/(s + 1)} + ?1{(13s ? 25)/((s ? 1)2 + (√3)2)}]

= (1/7)[e^(-t) + 13e^(t)cos(√(3)t) ? 4√(3)e^(t)sin(√(3)t)]

#1

Use partial fraction decomposition:

A/(s+1) + (Bs+C)/(s2-2s+4) = (2s2 - 2s - 3) / ((s + 1)(s2 - 2s + 4))

Multiply both sides by (s + 1)(s2 - 2s + 4)

A(s2-2s+4) + (Bs+C)(s+1) = 2s2 - 2s - 3

As2 - 2As + 4A + Bs2 + Bs + Cs + C = 2s2 - 2s - 3

(A + B)s2 + (-2A + B + C)s + (4A + C) = 2s2 - 2s - 3

Matching coefficients, we get:

A + B = 2

-2A + B + C = -2

4A + C = -3

Solving, we get A = 1/7, B = 13/7, C = -25/7

(2s2 - 2s - 3) / ((s + 1)(s2 - 2s + 4))

= (1/7)/(s+1) + (13/7 s - 25/7)/(s2-2s+4)

= 1/7 [1/(s+1) + (13s-13)/((s-1)2+(√3)2) - 12/((s-1)2+(√3)2)

= 1/7 [1/(s+1) + 13 (s-1)/((s-1)2+(√3)2) - (4√3) √3/((s-1)2+(√3)2)

Use table of Laplace transforms:

http://www.intmath.com/laplace-transform…

1/7 [e^(-t) + 13 e^(t) cos(√3 t) - 4√3 e^(t) sin(√3 t)]

Mαthmφm

#2