General solution to 2nd order DE

In summary, your initial conditions are x = 1 and t = 0, which gives you x' = 0 and t = 1. You substituted these values into your solution to get x(t) = e-t(Acos(4t) + Bsin(4t)).
  • #1
Dissonance in E
71
0
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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  • #2
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.
 
  • #3
Nice, thanks!
 
  • #4
What do I do if I am asked to write a solution that satisfies the conditions
x(0) = 1
x´(0) =0
 
  • #5
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.
 
  • #6
Cool, I got it now, thx.
 

What is a general solution to a 2nd order differential equation?

A general solution to a 2nd order differential equation is an equation that contains a constant term and can be used to find any particular solution to the equation by substituting specific values for the constants.

How do you find the general solution to a 2nd order differential equation?

To find the general solution to a 2nd order differential equation, you must first solve the equation by separating the variables and integrating. Then, you can add a constant term to the solution to get the general solution.

Can there be multiple general solutions to a 2nd order differential equation?

Yes, there can be multiple general solutions to a 2nd order differential equation. Each general solution will have a different constant term, which will result in a different particular solution when substituted with specific values.

What is the difference between a general solution and a particular solution to a 2nd order differential equation?

A general solution to a 2nd order differential equation contains a constant term and can be used to find any particular solution by substituting specific values for the constants. A particular solution is a specific solution to the equation that is obtained by substituting values for the constants in the general solution.

Can the general solution to a 2nd order differential equation be used to find all solutions to the equation?

No, the general solution to a 2nd order differential equation can only be used to find a particular solution by substituting specific values for the constants. To find all solutions to the equation, you would need to use initial conditions and the general solution to find the unique particular solution.

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