Inverse Laplace Transform, can't use partial fractions

[_diego]

Hi, this is my first post, and I'm not sure if I'm posting this in the right place.

Homework Statement

I have a 2nd order diff. equation:
3y''(t) + 2y'(t) + 5y(t) = 3
with initial values: y'(0) = 1, y(0) = 0

Homework Equations

After using Laplace transform I get:

Y(s) = (3 + 3s) / (s*[3s² +2s +5])

I believe it is correct, but even if it's not, what I'm interested in is how to solve this particular inverse laplace transform where I can't use partial fractions due to 3s² +2s +5 not having any roots.

The Attempt at a Solution

I'm kind of stuck here, but maybe I could split the fraction like:

Y(s) = 3 / (s*[3s² +2s +5]) + 3 / ([3s² +2s +5])

but the problem would remain.

Thanks

 

[gneill]

You could split it as

A/s + (Bs + C)/(3s² + 2s + 5)

 

[RTW69]

I would us an online solver such as WolframAlpha type in:

DSolve[{3y''[t]+2y'[t]+5y[t]==3,y'[0]==1,y[0]==0},y[t],t] and enter

 

[_diego]

Re-reading my post I think I didn't explain my problem correctly. My problem is in calculating the inverse transform of such rational function.
I could split it like that, but then I would have no idea on how to solve:
L^-1[ (Bs + C)/(3s² + 2s + 5) ]

Since I'm studying for a final exam, I really can't use a solver, but thanks.

 

[RTW69]

Thanks for the clarification. Looking at the WolframAlpha solution can still be helpful. Their solution looks very similar to a homogeneous second order diff. eqn with complex roots. Might be a place to start.

 

[Ray Vickson]

Hi, this is my first post, and I'm not sure if I'm posting this in the right place.

Homework Statement

I have a 2nd order diff. equation:
3y''(t) + 2y'(t) + 5y(t) = 3
with initial values: y'(0) = 1, y(0) = 0

Homework Equations

After using Laplace transform I get:

Y(s) = (3 + 3s) / (s*[3s² +2s +5])

I believe it is correct, but even if it's not, what I'm interested in is how to solve this particular inverse laplace transform where I can't use partial fractions due to 3s² +2s +5 not having any roots.

The Attempt at a Solution

I'm kind of stuck here, but maybe I could split the fraction like:

Y(s) = 3 / (s*[3s² +2s +5]) + 3 / ([3s² +2s +5])

but the problem would remain.

Thanks

Of course you can use partial fractions. The function 3s^2 + 2s + 5 has the form 3*(s-r1)*(s-r2), where r1 and r2 are the two roots of the quadratic; they are complex: r1 = -(1/3) + (1/3)*sqrt(14)*I and r2 = -(1/3) - (1/3)*sqrt(14)*I, where I = sqrt(-1). Then 1/[(s-r1)(s-r2)] = [1/(s-r1) - 1/(s-r2)]/(r1 - r2). The inverse Laplace transform of 1/(3s^2+2s+5) is [(1/3)/(r1-r2)]*[exp(r1*t) - exp(r2*t)] = (1/3)*exp(-t/3)*[(2i)sin(w*t)]/(2i*sqrt(14)/3) = (3/14)*exp(-t/3)*sin(wt), where w = sqrt(14)/3.

R.G. Vickson