Laplace Transform solution for 2nd order differential equation

khnbaba · May 11, 2014

Homework Statement

d^2x/dt^2 - 3 dx/dt + 2x= 2e^3t

give that at t=0, x=5, and dx/dt=7

Homework Equations

i can't figure out how to derive the values of A, B, and C from the attempted equation solution. please help me out here. thanks

The Attempt at a Solution

Handwritten working solving a second-order differential equation with Laplace transforms, ending with F(s) = [2/(s-3) + 5s - 8] / (s^2 - 3s + 2), with initial conditions x(0)=5 and x'(0)=7 substituted.

$Handwritten partial fraction decomposition of F(s) = (5s^2 - 23s + 26) / ((s-1)(s-2)(s-3)) with A/(s-1) + B/(s-2) + C/(s-3), and the step substituting s = 0 to find the coefficients.$

Handwritten working on paper showing the equation 2b = 6A + 3B + 2C struck through, then the note 'Setting coefficients equal and solving for the constants gives:' followed by three equations: A + B + C = 5, -5A - 4B - 3C = -23, and 6A + 3B + 2C = 26, labelled s^2, s^1 and s^0.

SteamKing · May 11, 2014

You've got three simultaneous equations in three unknowns, A, B, and C.

You can use elimination, Cramer's rule, guessing, whatever to solve for A, B, and C.

You've done good work up to this point. I'm surprised this point has stumped you.

khnbaba · May 11, 2014

thanks a lot steamKing you are being a big help, i went through cramer's rule examples and i think its going to solve my problem, i am going to try it now. plus i will learn some thing new :). i really appreciate your help. once again thanks

khnbaba · May 11, 2014

Handwritten working solving three simultaneous equations in A, B, C using Cramer's rule, with determinants D, DA, DB, DC calculated and final values A = 209, B = 0, C = 1.

$Handwritten working applying partial fractions and inverse Laplace transforms: (26 + 5s^2 - 23s)/((s-1)(s-2)(s-3)) = A/(s-1) + B/(s-2) + C/(s-3), giving F(s) = 209/(s-1) + 1/(s-3), then y(t) = e^(209t) + e^(3t).$

SteamKing can you have a look to the final answer. thanks

SteamKing · May 11, 2014

For the second and third equations in your original system, for some reason you have the coefficients of A and C switched around.

Check your arithmetic in calculating the coefficients from the steps above that point, specifically where you let s = 0. Is the product of two negative numbers another negative number?

khnbaba · May 11, 2014

are you talking about equation 2, and eq 3 in the beginning?
and s=0 was just a supposition and the arithmetic below isn't part of the question which i realize now the product of two negative numbers should be positive not negative.

SteamKing · May 11, 2014

khnbaba said:

are you talking about equation 2, and eq 3 in the beginning?
and s=0 was just a supposition and the arithmetic below isn't part of the question which i realize now the product of two negative numbers should be positive not negative.

Yes. It appeared from your work that the calculations where s = 0 is where you started to calculate the coefficients of A, B, and C.

khnbaba · May 11, 2014

thanks a lot for your time and efforts.

Laplace Transform solution for 2nd order differential equation

Homework Statement

Homework Equations

The Attempt at a Solution

Discussion

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