- #1

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I start by rewriting the formula as such:

dv/dt=k*dx/dt*x

I am confused now though because I now have dv,dt, and dx. I assume I need to get everything in terms of dx and dt. Can I use seperation of variables or do I need another technique?

- Thread starter infraray
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- #1

- 23

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I start by rewriting the formula as such:

dv/dt=k*dx/dt*x

I am confused now though because I now have dv,dt, and dx. I assume I need to get everything in terms of dx and dt. Can I use seperation of variables or do I need another technique?

- #2

quasar987

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If so, you got F=kvx, which can be rewriten as

[tex]F = k\frac{dx}{dt}x[/tex]

and F = ma, which can be rewritten as

[tex]F = m\frac{d^2x}{dt^2}[/tex]

Combining the two leads to

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

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Now is this where you have trouble? [itex]kx=mx^{2}[/itex] has solutions x=0 and x=k/m. What does this tell you about the differential equation?

Last edited:

- #4

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

Since it doesn't appear to be able to be put into a 1st order equation I assume I am out of luck with seperation, or am I?

Sorry, but I'm not sure what x=0, x=k/m tells me about the equation. Are they linear dependent? The lighter the object the further it travels?

It's been over 2 years since diff eq and my head is a mess right now. I feel helpless. My teacher keeps saying this is easy, but when you've been away from it for a while, it is very hard. It's not like riding a bike that's for sure.

- #5

HallsofIvy

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[tex]\frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}= v\frac{dv}{dx}[/tex]

to eliminate t from the equation.

[tex]m\frac{dv}{dt}= mv\frac{dv}{dx}= F= kxv[/tex]

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