General solution to 2nd order DE

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Homework Help Overview

The discussion revolves around finding the general solution to a second-order differential equation, specifically x'' - 2x' + 5x = 0. Participants are exploring the characteristic polynomial and the implications of complex roots in the context of differential equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss forming the characteristic polynomial and solving for roots, with some questioning the correctness of their general solution format. There is also a query regarding how to apply initial conditions to find specific constants in the solution.

Discussion Status

Some participants have provided clarifications on the general solution form and the application of initial conditions. There is ongoing exploration of how to derive constants from the given conditions, indicating a productive direction in the discussion.

Contextual Notes

Initial conditions x(0) = 1 and x'(0) = 0 are introduced, prompting further discussion on how to utilize these conditions to solve for constants in the general solution.

Dissonance in E
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Hi, I am just looking for clarification whether or not I am doing these problems correctly, here's an example:

Homework Statement


Find the general solution to
x'' -2x' + 5x = 0


Homework Equations


Charasteristic polynomial.
Quadratic equation.
General solution form for complex roots.

The Attempt at a Solution


form the char polynomial:
m^2 -2m +5 = 0

Solve for roots:
(-2 (+-)SQRT(2^2 - 4(1)(5)))/2
-1+SQRT(-16)
-1-SQRT(-16)
m1 = -1+4i
m2 = -1-4i

Insert into general solution form:

y = (e^ax)(Acoswx + Bsinwx)
y = (e^-1x)(Acos4ix + Bsin4ix)

General solution: (?)
y = (e^-x)(Acos4ix + Bsin4ix)
 
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Dissonance in E said:
Hi, I am just looking for clarification whether or not I am doing these problems correctly, here's an example:

Homework Statement


Find the general solution to
x'' -2x' + 5x = 0


Homework Equations


Charasteristic polynomial.
Quadratic equation.
General solution form for complex roots.

The Attempt at a Solution


form the char polynomial:
m^2 -2m +5 = 0

Solve for roots:
(-2 (+-)SQRT(2^2 - 4(1)(5)))/2
-1+SQRT(-16)
-1-SQRT(-16)
m1 = -1+4i
m2 = -1-4i

Insert into general solution form:

y = (e^ax)(Acoswx + Bsinwx)
y = (e^-1x)(Acos4ix + Bsin4ix)

General solution: (?)
y = (e^-x)(Acos4ix + Bsin4ix)

Almost!
Your general solution is
y = e-x(Acos(4x) + Bsin(4x))

You can verify that this is the solution by calculating y' and y'' and substituting them into your differential equation.
 
Nice, thanks!
 
What do I do if I am asked to write a solution that satisfies the conditions
x(0) = 1
x´(0) =0
 
Your solution is actually x(t) = e-t(Acos(4t) + Bsin(4t)). I hadn't noticed earlier that you had switched your solution to be y as a function of x.

Use your initial conditions to substitute x = 1 and t = 0 into your solution, and x' = 0 and t = 1 into the derivative of your solution. That will give you two equations in two unknowns, so you should be able to solve for the constants A and B.
 
Cool, I got it now, thx.
 

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