- #1

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I used seperation of variables to get

x/dx=(1-4v^2)/3v dv

I'm not sure if thats even right.

But if it is right, how do I integrate that?

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- Thread starter EP
- Start date

- #1

- 76

- 0

I used seperation of variables to get

x/dx=(1-4v^2)/3v dv

I'm not sure if thats even right.

But if it is right, how do I integrate that?

- #2

Integral

Staff Emeritus

Science Advisor

Gold Member

- 7,212

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Your Separation of variables looks good, now a simple substitution finishes the job.

- #3

- 391

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You made a slight mistake with the variable seperation--- you can't have dx in the denominator! (Atleast, I've learnt not to do it)EP said:

I used seperation of variables to get

x/dx=(1-4v^2)/3v dv

I'm not sure if thats even right.

But if it is right, how do I integrate that?

You should have done:

xdv = (1-4v^2)/3v dx

dx/x = 3v/(1-4v^2) dv

Integration from there is pretty simple.... apply your initial conditions. :)

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