# Homework Help: Inverse laplace transform

1. Mar 5, 2012

### aaaa202

Using the laplace transform, find the solution to the differential equation:

y'' + y' + y = 0 , y(0)=0, y'(0)=1

Using the laplace transform and its properties I end up with:

f(s) = 1/(s2+s+1)

How can I find the inverse of this/ does anyone know the inverse of it?

Setting y=eax I got a characteristic equation of a2+a+1=0, which has a complex solution, so I suspec that the inverse above should be a combination of both a sine, cosine and an exponential..

2. Mar 5, 2012

### Ray Vickson

Factor the denominator, then use partial fractions.

RGV

3. Mar 5, 2012

### aaaa202

hmm okay so factoring the denominator I get:

f(s) = 1/((s+½+i√(3)/2)(s+½-i√(3)/2))

What do I then do? Sorry, but I don't know the term partial fractions :(

4. Mar 5, 2012

### Ray Vickson

Google "partial fractions", or look in any calculus textbook.

RGV

5. Mar 5, 2012

### vela

Staff Emeritus
You could also complete the square in the denominator and then use the tables, but you should know how to invert the transform using either method.

6. Mar 5, 2012

### aaaa202

hmm using partial fractions I get:

1/(s2+s+1) = -i√(3)/(s+½+i√(3)/2) + i√(3)/(s+½-i√(3)/2)

But I don't see how that gets me anywhere.

Instead I figured you might want to write it as:

1/((s+½)2+3/4)

But can't find a specific transform looking like that.

The issue is that I have already obtained the solution using the characteristic polynomial, which got the solution:

e^(½x)(cos(√(3)/2 *x) + 1/√(3) * sin(√(3)/2 * x))

But I can't find any transform tables with e^ax(sin(bx) or with cos(bx)

Last edited: Mar 5, 2012
7. Mar 5, 2012

### sunjin09

If you write your solution all in terms of complex exponentials, you'll probably find correspondence in tables.

8. Mar 5, 2012

### aaaa202

No because I'm not considering complex solutions (there's no i in front of sine). Hmm this is getting too tedious anyways, Ima skip this assignment. After all I know what to do, and who cares I can't get it to fit completely.

9. Mar 5, 2012

### Ray Vickson

Well, you could work these out on your own; they are not difficult (but are a bit lengthy before the final result).

Anyway, you have something of the form $\frac{c_1}{s+a_1} + \frac{c_2}{s + a_2},$ for constants $c_1, c_2, a_1, a_2$ (which happen to be complex numbers, but that does not matter). Just invert each term separately. In the end, you will be able to combine the results to get a purely real result.

RGV