Did I Apply Laplace Transforms Correctly?

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SUMMARY

The discussion revolves around the application of Laplace Transforms to solve the differential equation $y'' - 5y' + 6y = 1$ with initial conditions $y(0) = 1$ and $y'(0) = 0$. The user correctly transforms the equation into the Laplace domain, resulting in the expression $Y = \frac{1+s^2-5s}{s^3-5s^2+6s}$. They also set up the partial fraction decomposition as $1+s^2-5s = \frac{A}{s} + \frac{B}{s-2} + \frac{C}{s-3}$, confirming the correctness of their approach before concluding that they resolved their confusion.

PREREQUISITES
  • Understanding of differential equations, specifically second-order linear equations.
  • Familiarity with Laplace Transforms and their properties.
  • Knowledge of partial fraction decomposition techniques.
  • Basic algebraic manipulation skills for handling polynomial expressions.
NEXT STEPS
  • Study the application of Laplace Transforms in solving higher-order differential equations.
  • Learn about the inverse Laplace Transform and its techniques.
  • Explore the method of undetermined coefficients for solving differential equations.
  • Investigate the stability analysis of solutions to differential equations using Laplace Transforms.
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Students and professionals in mathematics, engineering, and physics who are working with differential equations and wish to deepen their understanding of Laplace Transforms and their applications.

shamieh
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Solev by Laplace Transforms

$y'' - 5y' + 6y = 1$ $y(0) = 1$, $y'(0) = 0$So I am getting stuck. Here's my work

$s^2Y - 5sY + 6Y = \frac{1}{s} + s - 5$

multiplied through by $s$ to get

$s^3Y - 5s^2Y + 6sY = 1 + s^2 - 5s$

so:

$Y = \frac{1+s^2-5s}{s^3-5s^2+6s}$

so: $1+s^2-5s = \frac{A}{s} + \frac{B}{s-2} + \frac{C}{s-3}$

so: is it correct to say $1+s^2-5s = A(s-2)(s-3) + Bs(s-3) + Cs(s-2)$
 
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snvm i figured it out lol
 

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