Laplace Transform solution for 2nd order differential equation

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SUMMARY

The discussion focuses on solving a second-order differential equation using the Laplace Transform method. The equation presented is d²x/dt² - 3 dx/dt + 2x = 2e³t, with initial conditions x(0) = 5 and dx/dt(0) = 7. Participants emphasize the importance of correctly applying Cramer's rule to derive the coefficients A, B, and C from the system of equations generated by the Laplace Transform. Additionally, they highlight the need for careful arithmetic, particularly when evaluating coefficients at s = 0.

PREREQUISITES
  • Understanding of Laplace Transform techniques
  • Familiarity with second-order differential equations
  • Knowledge of Cramer's rule for solving systems of equations
  • Basic arithmetic and algebra skills
NEXT STEPS
  • Study the application of Laplace Transforms in solving differential equations
  • Practice using Cramer's rule with various examples
  • Review the properties of Laplace Transforms, particularly initial value problems
  • Explore common pitfalls in arithmetic when solving for coefficients in differential equations
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Students studying differential equations, educators teaching advanced mathematics, and anyone seeking to improve their problem-solving skills in applied mathematics.

khnbaba
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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


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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.
 
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
 
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SteamKing can you have a look to the final answer. thanks
 
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?
 
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.
 
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.
 
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thanks a lot for your time and efforts.
 

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