Fraction decomposition for inverse Laplace

[nateja]

Homework Statement

Find the solution of the givien initial value problem and draw its graph
y''+2y'+2y = δ(t-π) y(0) = 1, y'(0) = 0

Homework Equations

A Laplace transform chart would be very useful

The Attempt at a Solution

I chose to solve the equation with Laplace transforms. I took the laplace of the whole equation and then solved for the L(y):

L(y) = (e^(-πs)+s+2)/(s^2+2s+2)
I double checked to make sure I didn't make any errors in my Laplace calculation and I did the algerbra correctly. So I have the right equation. However, my issue is now decomposing this equation - how do you do that?

You can't factor out the denominator? and I can't just set the numerator to As+B can I? The e^s term would make that useless...

Any ideas?

 

[nateja]

and just to explain, π is pi and δ is delta for dirac delta function

 

[nateja]

I tried to use complete the square on the denominator to see if I could break it up more. I got s = -1 + i and s = -1 - i... so I don't know what to do with that at all. I'll keep trying though...

 

[Mark44]

Homework Statement

Find the solution of the givien initial value problem and draw its graph
y''+2y'+2y = δ(t-π) y(0) = 1, y'(0) = 0

Homework Equations

A Laplace transform chart would be very useful

The Attempt at a Solution

I chose to solve the equation with Laplace transforms. I took the laplace of the whole equation and then solved for the L(y):

L(y) = (e^(-πs)+s+2)/(s^2+2s+2)
I double checked to make sure I didn't make any errors in my Laplace calculation and I did the algerbra correctly. So I have the right equation. However, my issue is now decomposing this equation - how do you do that?

You can't factor out the denominator? and I can't just set the numerator to As+B can I? The e^s term would make that useless...

Any ideas?

If you complete the square in the denominator, you'll get s² + 2s + 1 + 1 = (s + 1)² + 1.

 

[Ray Vickson]

Homework Statement

Find the solution of the givien initial value problem and draw its graph
y''+2y'+2y = δ(t-π) y(0) = 1, y'(0) = 0

Homework Equations

A Laplace transform chart would be very useful

The Attempt at a Solution

I chose to solve the equation with Laplace transforms. I took the laplace of the whole equation and then solved for the L(y):

L(y) = (e^(-πs)+s+2)/(s^2+2s+2)
I double checked to make sure I didn't make any errors in my Laplace calculation and I did the algerbra correctly. So I have the right equation. However, my issue is now decomposing this equation - how do you do that?

You can't factor out the denominator? and I can't just set the numerator to As+B can I? The e^s term would make that useless...

Any ideas?

L(y)(s) = exp(-πs)/(s^2+2s+2) + (s+2)/(s^2+2s+2). Can you find the inverse LT of 1/(s^2+2s+2)? If so, getting the inverse LT of exp(-πs)/(s^2+2s+2) is easy, just using some standard results about Laplace transforms. You can look them up for yourself.

 

[nateja]

L(y)(s) = exp(-πs)/(s^2+2s+2) + (s+2)/(s^2+2s+2).

I got this. I also noticed my mistake in completing the square and solved that. I got (s+1)^2 +1 for the denominator now.

So I should get the inverse laplace as u_π(t)*t*e^(-t) + ?

I am getting (s+2)/((s+1)^2 + 1) which does not have an inverse laplace in the chart. The nearest thing is laplace of e^(at)*cos(bt) which is not exactly equivalent because the 'a' term doesn't match.

 

[nateja]

Also,

Laplace inverse of s/((s+1)^2+1) + 2 * 1/((s+1)^2+1) does not work out either.

The first term is close to the laplace of e^(at)*cos(bt); however, it is not exact again. The second term will be 2*e^(-t)*sin(t)

I can't see any way to tinker with the algerbra without adding on new terms and creating more inverse laplaces...

Would adding (1-1)/((s+1)^2+1) to y give you y = u_π(t)*t*e^(-t) + e^(-t)cos(t) + e^(-t)*sin(t)

I'm pretty sure adding 0 is legal. Does that check out with you guys?