Differential Equation Problem?

Feldoh
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Homework Statement


In 1692, Johann Bernoulli was teaching the Marquis de l'Hopital calculus in Paris. Solve the following problem, which is similar to the one they did. What is the equation of the curve which has subtangent equal to twice its abscissa.

Homework Equations


None


The Attempt at a Solution


Honestly I'm not really sure where to start. Heck, I'm not even seeing how it's part of the differential equations chapter in my math book. Could anyone just me a push in te right direction?
 
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The subtangent is just f(x)/f'(x), isn't it? What's twice the abscissa? Looks like an ODE to me.
 
The subtangent is twice the abscissa (which is like a distance on the x-axis?) so...

2f(x)/f'(x) = x?
 
Looks more to me like f(x)/f'(x)=2x.
 
Dick said:
Looks more to me like f(x)/f'(x)=2x.

Yeah I just fail at reading XD

Thanks. But I do have one more question what exactly is a subtangent?
 
Last edited:
Feldoh said:
Yeah I just fail at reading XD

Thanks. But I do have one more question what exactly is a subtangent?

I thought YOU knew! I had to google it. http://en.wikipedia.org/wiki/Subtangent
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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