General Solution / Differential Equation

[emergentecon]

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?

 

[Dick]

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.

 

[emergentecon]

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?

 

[pasmith]

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

Five times an arbitrary constant is an arbitrary constant.

 

[SteamKing]

Substitute your solution into the original ODE and see if it satisfies that equation. Do likewise with Mathematica's result.

 

[emergentecon]

Five times an arbitrary constant is an arbitrary constant.

Hehe, fair enough.

Thanks!