How Do You Perform an Inverse Laplace Transform on 1/(s^2+s+1)?

[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=e^ax 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..

 

[Ray Vickson]

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=e^ax 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..

Factor the denominator, then use partial fractions.

RGV

 

[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 :(

 

[Ray Vickson]

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 :(

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

RGV

 

[vela]

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.

 

[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)

 

[sunjin09]

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)

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

 

[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.

 

[Ray Vickson]

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)

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