- #1

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[tex]\dot{x} = Hx_0 \left( \sqrt{B} \frac{x_0}{x} + \sqrt{1-A-B} \right)[/tex]

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- Thread starter Logarythmic
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- #1

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[tex]\dot{x} = Hx_0 \left( \sqrt{B} \frac{x_0}{x} + \sqrt{1-A-B} \right)[/tex]

- #2

mjsd

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[tex]\dot{x} = Hx_0 \left( \sqrt{B} \frac{x_0}{x} + \sqrt{1-A-B} \right)[/tex]

what is what? in any case, try integrating factor method

- #3

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It's equivalent to solving

[tex]y \frac{dy}{dx} + ay + b = 0[/tex]

[tex]y \frac{dy}{dx} + ay + b = 0[/tex]

- #4

HallsofIvy

Science Advisor

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We need more information. Are you saying that A, B, H, and x_{0} are constants? Is so, simplify by letting

[tex]U= \frac{H\sqrt{B}x_0^2}{x}[/tex]

and

[tex]V= Hx_0\sqrt{1- A- B}[/itex]

So your equation becomes

[tex]\frac{dx}{dt}= \frac{U}{x}+ V= \frac{U+ Vx}{x}[/tex]

[tex]\frac{xdx}{U+ Vx}= dt[/itex]

That's easy to integrate.

[tex]U= \frac{H\sqrt{B}x_0^2}{x}[/tex]

and

[tex]V= Hx_0\sqrt{1- A- B}[/itex]

So your equation becomes

[tex]\frac{dx}{dt}= \frac{U}{x}+ V= \frac{U+ Vx}{x}[/tex]

[tex]\frac{xdx}{U+ Vx}= dt[/itex]

That's easy to integrate.

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- #5

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[tex]t= \frac{x}{V} -\frac{U}{V^2}ln(Vx+U)[/tex]

and trying to solve this for x is rather difficult?

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