General Solution / Differential Equation

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Homework Help Overview

The discussion revolves around finding the general solution x(t) to a differential equation given by dx/dt = 2t/5x. Participants are exploring the implications of their solutions and the behavior of arbitrary constants in the context of differential equations.

Discussion Character

  • Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants present their attempts at solving the differential equation and express confusion regarding the appearance of a constant factor in the solution provided by Mathematica. They question the equivalence of arbitrary constants and whether the notation implies a different meaning.

Discussion Status

The discussion is ongoing, with participants clarifying their understanding of arbitrary constants and exploring the validity of their solutions. Some guidance has been offered regarding the nature of constants, but no consensus has been reached on the implications of the differing constants.

Contextual Notes

Participants are working under the constraints of homework rules, focusing on the interpretation of their solutions and the definitions of constants in the context of differential equations.

emergentecon
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Homework Statement


Find the general solution x(t) to the following differential equation:

dx/dt = 2t/5x

Homework Equations



dx/dt = 2t/5x

The Attempt at a Solution


My solution is:

∫5xdx = ∫2tdt
(5/2)x^2 = t^2 + C
x^2 = (2/5)(t^2 + C)
x = +-√[(2/5)(t^2 + C)]

However, when I put the problem in Mathematica, I get:

x = +-√(2/5)√(x^2 + 5C)

I don't see why it gets 5C as opposed to simply C?
 
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emergentecon said:

Homework Statement


Find the general solution x(t) to the following differential equation:

dx/dt = 2t/5x

Homework Equations



dx/dt = 2t/5x

The Attempt at a Solution


My solution is:

∫5xdx = ∫2tdt
(5/2)x^2 = t^2 + C
x^2 = (2/5)(t^2 + C)
x = +-√[(2/5)(t^2 + C)]

However, when I put the problem in Mathematica, I get:

x = +-√(2/5)√(x^2 + 5C)

I don't see why it gets 5C as opposed to simply C?

No reason. C is an arbitrary constant and 5C is also an arbitrary constant. There's really no difference. Both are correct.
 
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Dick said:
No reason. C is an arbitrary constant and 5C is also an arbitrary constant. There's really no difference. Both are correct.

I understood it to mean 5 * C . . . is that not correct?
 
emergentecon said:
I understood it to mean 5 * C . . . is that not correct?

Five times an arbitrary constant is an arbitrary constant.
 
Substitute your solution into the original ODE and see if it satisfies that equation. Do likewise with Mathematica's result.
 
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pasmith said:
Five times an arbitrary constant is an arbitrary constant.

Hehe, fair enough.

Thanks!
 

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