CMPSC 200 10 B Analytical Solution to an ODE II
Автор: Joseph Mahoney
Загружено: 2015-07-27
Просмотров: 102
Описание:
Analytical Solution to a Second-Order ODE with Initial Conditions | x'' + 5x' + 6x = 0, x(0)=1, x'(0)=-1
This video walks through a full hand-solution of a second-order, linear, homogeneous ordinary differential equation with constant coefficients. You will see how the characteristic equation method turns the differential equation into an algebra problem, then how two initial conditions determine the two constants in the general solution.
ODE (time-domain model):
x''(t) + 5x'(t) + 6x(t) = 0
Initial conditions:
x(0) = 1
x'(0) = -1
What you will learn in this worked example:
We form the characteristic equation r^2 + 5r + 6 = 0, factor it to find the roots r = -3 and r = -2, and write the general solution
x(t) = A e^(-3t) + B e^(-2t).
Then we apply x(0) and x'(0) to build a 2-equation system and solve for A and B, giving:
x(t) = e^(-3t) - 2 e^(-2t).
Quick check: substituting x(t) back into x'' + 5x' + 6x confirms the left-hand side equals 0, and both initial conditions are satisfied.
More ODE resources and videos (course page):
(https://sites.google.com/view/jmm-mat...)
Related videos on this channel (good next steps):
Analytical Solution to an ODE I (separable first-order example): ( • CMPSC 200 10 A Analytical Solution to an O... )
Solving ODEs Numerically (state space setup and solver workflow): ( • CMPSC 200 10 C Solving ODEs numerically )
Using ODE45 in MATLAB: ( • CMPSC 200 10 F Using ODE45: how to use ODE... )
ODE45 Example I (worked numerical solution): ( • CMPSC 200 10 G ODE45 Example I: solving ou... )
Introduction to System Dynamics (how ODE models show up in engineering): ( • ME 357 00 A Introduction to System Dynamics )
2nd-Order Systems Introduction (system dynamics connection): ( • ME 357 13 A1 2nd-Order Systems Introduction )
#DifferentialEquations #ODE #EngineeringMath #SystemDynamics #MATLAB #ControlSystems
Timestamps
0:00 Problem statement and initial conditions
0:47 Build the characteristic equation
1:05 Factor to find the roots
1:23 Write the general exponential solution
2:08 Apply x(0)=1 to get equation 1
2:38 Differentiate and apply x'(0)=-1 to get equation 2
3:18 Solve for A and B
3:43 Final x(t) and initial-condition checks
4:07 Substitute back into the ODE to verify
Повторяем попытку...
Доступные форматы для скачивания:
Скачать видео
-
Информация по загрузке: