Laplace transform applied to a differential equation

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SUMMARY

The discussion focuses on solving the differential equation \(\frac{dy}{dt} - y = 1\) with the initial condition \(y(0) = 0\) using the Laplace transform. The correct application of the Laplace transform yields \(Y(s) = \frac{1}{s^3} + \frac{1}{s^2}\), leading to the solution \(y(t) = \frac{1}{2}t^2 + t\). An error was identified in the transformation of \(\mathcal{L}\{y\}\), which was crucial for obtaining the correct solution. The resolution of this error clarified the discrepancy with the solution obtained through separation of variables.

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  • Understanding of differential equations, specifically first-order linear equations.
  • Familiarity with the Laplace transform and its properties.
  • Knowledge of initial value problems and their solutions.
  • Ability to perform inverse Laplace transforms.
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  • Study the properties of the Laplace transform in detail.
  • Learn about solving initial value problems using Laplace transforms.
  • Explore the method of separation of variables for differential equations.
  • Practice inverse Laplace transforms with various functions.
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Students and professionals in mathematics, engineering, and physics who are solving differential equations and applying the Laplace transform in their work.

Lord Anoobis
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Homework Statement


Solve ##\frac{dy}{dt} -y = 1, y(0) = 0## using the laplace transform

Homework Equations

The Attempt at a Solution


##\mathcal{L}\big\{\frac{dy}{dt}\big\} - \mathcal{L}\big\{y\big\} = \mathcal{L}\big\{1\big\}##

##sY(s) - y(0) - \frac{1}{s^2} = \frac{1}{s}##

##Y(s) =\frac{1}{s^3} + \frac{1}{s^2} ##

Applying the inverse Laplace leads to:

##y(t) = \frac{1}{2}t^2 + t##

Which is nothing like the answer obtained with separation of variables. What have I done wrong here?
 
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Okay, I see the trouble. Made an error with ##\mathcal{L}\big\{y\big\}##. Problem solved
 

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