Equivalence of Two Forms of Solutions to Second Order ODEs

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In summary, the two solutions to a second order ODE are the linear combination of two solutions and the phase-shifted sin/cos function. The first solution can be verified from the characteristic equation for the differential equation, and the second solution can be verified from the equation for a harmonic oscillator. However, the second solution does not satisfy the equation.
  • #1
nickmai123
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Homework Statement


I was curious if anyone could help me prove the equivalence between the two forms of solutions to second order ODEs, one being the linear combination of two solutions and the other being the phase-shifted sin/cos function.


Homework Equations


[tex]\frac{d^{2}x}{dt^{2}}+\frac{k}{m}x=0[/tex]

[tex]x(t)=Asin(\omega t+\phi)[/tex]

[tex]x(t)=C_{1}cos(\sqrt{\frac{k}{m}}t)+C_{2}sin(\sqrt{\frac{k}{m}}t[/tex]

[tex]A[/tex] and [tex]\phi[/tex] should be treated as constants. I know that this is the differential equation for a harmonic oscillator, but I figured since it's more of how the mathematics behind it work, it belongs in this forum. If the moderator believes it should be moved, then by all means please move it.

The Attempt at a Solution


I have no idea where to start the proof for this. The second solution I wrote can be simply verified from the characteristic equation for the differential equation.

Thanks.

EDIT: I found my error. Oops. I wrote it as r^2 + (k/m)r. Thats what I get for doing math at 3AM, lol.
 
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  • #2
You're missing the second exponential in the 2nd equation.
 
  • #3
Pengwuino said:
You're missing the second exponential in the 2nd equation.

Hm? Which exponential? I got roots of 0 and -k/m for that differential equation's characteristic equation. I did forget the negative sign to the -k/m though.
 
  • #4
Check the roots again, the characteristic equation should be [tex]\lambda ^2 + \frac{k}{m} = 0[/tex]. Assuming k>0, you have imaginary roots giving a positive and negative exponential solution.
 
  • #5
I don't think the second solution satisfies the equation.

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  • #6
nickmai123 said:

Homework Statement


I was curious if anyone could help me prove the equivalence between the two forms of solutions to second order ODEs, one being the linear combination of two solutions and the other being the phase-shifted sin/cos function.


Homework Equations


[tex]\frac{d^{2}x}{dt^{2}}+\frac{k}{m}x=0[/tex]

[tex]x(t)=Asin(\omega t+\phi)[/tex]

[tex]x(t)=C_{1}+C_{2}e^{-\frac{k}{m}t}[/tex]

[tex]A[/tex] and [tex]\phi[/tex] should be treated as constants. I know that this is the differential equation for a harmonic oscillator, but I figured since it's more of how the mathematics behind it work, it belongs in this forum. If the moderator believes it should be moved, then by all means please move it.

The Attempt at a Solution


I have no idea where to start the proof for this. The second solution I wrote can be simply verified from the characteristic equation for the differential equation.

Thanks.

nickmai123 said:
Hm? Which exponential? I got roots of 0 and -k/m for that differential equation's characteristic equation. I did forget the negative sign to the -k/m though.
Then you've either got the wrong characteristic equation or you have solved it incorrectly. Neither 0 nor -k/m satisfy the characteristic equation for this d.e.

Please show your work.
 
  • #7
HallsofIvy said:
Then you've either got the wrong characteristic equation or you have solved it incorrectly. Neither 0 nor -k/m satisfy the characteristic equation for this d.e.

Please show your work.

Got it. Thanks.
 
  • #8
Oh, that's not fair! Show your work so we can all laugh at your mistake and feel superior!

(Some silly little arithmetic or algebra mistake.)
 
  • #9
HallsofIvy said:
Oh, that's not fair! Show your work so we can all laugh at your mistake and feel superior!

(Some silly little arithmetic or algebra mistake.)

Haha no i just added an r for the term (k/m)x. I took DE my sr year in high school, so most of it's begun to leave me. I have a feeling it's going to come back bite me in the butt when I take circuits... sigh.
 

FAQ: Equivalence of Two Forms of Solutions to Second Order ODEs

What is a solution of a 2nd order differential equation?

A solution of a 2nd order differential equation is a function that satisfies the equation and its initial conditions. It is a way to represent the relationship between the dependent and independent variables in the equation.

What is the general form of a 2nd order differential equation?

The general form of a 2nd order differential equation is: $$a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x)$$where a, b, and c are constants and f(x) is a function of x.

How do you solve a 2nd order differential equation?

There are several methods for solving a 2nd order differential equation, including the method of undetermined coefficients, variation of parameters, and using Laplace transforms. The specific method used depends on the form of the equation and its initial conditions.

What are the types of solutions for a 2nd order differential equation?

The solutions of a 2nd order differential equation can be classified into three types: general solution, particular solution, and singular solution. The general solution represents the family of all possible solutions, the particular solution satisfies the initial conditions, and the singular solution is a special solution that cannot be obtained from the general solution.

What are the applications of 2nd order differential equations in science?

2nd order differential equations have many applications in science, particularly in the fields of physics, engineering, and chemistry. They are used to model various physical phenomena such as motion, heat transfer, and chemical reactions. They are also essential in understanding the behavior of complex systems and making predictions based on mathematical models.

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