Laplace Transforms and Differential Equations

Flashcards based on lecture notes on solving ordinary differential equations using Laplace transforms.

Laplace Transform Technique

A powerful tool for solving physical systems involving ODEs, particularly initial value problems, by reducing the ODE solution to an algebraic equation.

Step 1 in solving ODEs using Laplace Transform

Take the Laplace transform of both sides of the ODE and initial conditions to convert the differential equation into an algebraic equation in y(s).

Step 2 in solving ODEs using Laplace Transform

Solve the algebraic equation for y(s).

Step 3 in solving ODEs using Laplace Transform

The inverse transformation ʆ−1{y(s)} = y(t) is the required solution of the given problem.

Laplace Transform of first derivative

ʆ{f'(t)} = sF(s) - f(0)

Laplace Transform of second derivative

ʆ{f''(t)} = s^2F(s) - sf(0) - f'(0)

Partial Fraction Decomposition

Used to simplify algebraic expressions obtained after applying Laplace transforms, allowing for easier inverse transformation.

Solution to the differential equation y''' - 6y'' + 11y' - 6y = e^(4t) with initial conditions y(0) = y'(0) = y''(0) = 0

y(t) = -1/6 e^t + 1/2 e^(2t) - 1/2 e^(3t) + 1/6 e^(4t)

Solution to the differential equation dy/dt - y = 1, y(0) = 0

y(t) = -1 - t + e^t

Solution to the differential equation y'' + y = sint, y(0) = 1, y'(0) = -1

y(t) = cost - t/2 sint

Solution to the differential equation d^2y/dt^2 + 2dy/dt - 3y = e^(-t), y(0) = 1, y'(0) = 0

y(t) = 11/16 e^(-t) + t/4 e^(-t) + 5/16 e^(-3t)

Solution to the differential equation d^2y/dt^2 = 20e^(-t)cost, y(0) = 0 = y'(0)

y(t) = -5 + 3e^(-t) + 2e^(cost)- 4e^(-t)sint