Really quick differential question

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Lewis
If I am rewriting v in terms of x and t, I write it as dx/dt, but how about v^2?
 
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urnt? Are you talking about Reduction of order?
Like:

http://tutorial.math.lamar.edu/AllBrowsers/3401/ReductionofOrder_files/eq0021M.gif

http://tutorial.math.lamar.edu/AllBrowsers/3401/ReductionofOrder_files/eq0022M.gif

http://tutorial.math.lamar.edu/AllBrowsers/3401/ReductionofOrder_files/eq0023M.gif
 
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I don't know, my prof likes to ask questions on things he doesn't teach :\

I mean, in an equation like: F=-mv, you can rewrite it like m [(d^2 x)/(dt^2)] = -m dx/dt

But how would you rewrite F=-m(v^2) ?
 
(v)^2 = v*v
(\frac{dx}{dt})^2 = \frac{dx}{dt} * \frac{dx}{dt}

Let's say x=t^2+t^4 then v = \frac{dx}{dt}. So we then have F = mv^2 = m \times \frac{dx}{dt} \frac{dx}{dt} = 2t+4t^3 \times 2t+4t^3

assuming F = mv
 
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Okay, thanks a lot!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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