Solving ODE involving square of first derivative

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The discussion revolves around solving a non-linear ordinary differential equation (ODE) derived from a problem in classical mechanics. The equation involves the square of the first derivative and constants related to speed and length. A suggested approach involves rewriting the equation to isolate the first derivative and then integrating using a trigonometric substitution. The user acknowledges a mistake in typing the ODE but believes the proposed method remains applicable. Overall, the conversation highlights the challenges of solving non-linear ODEs and the importance of proper mathematical techniques.
tsw99
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Homework Statement



this is not from a math course, but from Gregory's classical mechanics book prob 2.10
it's easy to obtain the desired ODE
\dot{r}^{2}=\frac{u^{2}}{a^{2}}(\frac{U^{2}a^{2}}{a^{2}}-r^{2})
since it's non-linear, i have a difficult time to solve for r(t)
u, U and a are some constants with unit speed, speed and length resp.

Homework Equations





The Attempt at a Solution


all methods i know fail, I believe there is some trick that I am not aware of. great appreciate for any help:(
 
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The equation you have can be written as:
<br /> \dot{r}=\pm\frac{u}{a}\sqrt{U^{2}-r^{2}}<br />
Dividing and integrating shows that:
<br /> \int\frac{dr}{\sqrt{U^{2}-r^{2}}}=\pm\frac{u}{a}\int dt<br />
The integral can be solved by the substitution:
<br /> r=U\sin\alpha<br />
I will leave you to slog through the algerbra.
 
hunt_mat said:
The equation you have can be written as:
<br /> \dot{r}=\pm\frac{u}{a}\sqrt{U^{2}-r^{2}}<br />
Dividing and integrating shows that:
<br /> \int\frac{dr}{\sqrt{U^{2}-r^{2}}}=\pm\frac{u}{a}\int dt<br />
The integral can be solved by the substitution:
<br /> r=U\sin\alpha<br />
I will leave you to slog through the algerbra.

oh...thank you very much!
I think i need to brush up my math skills...edit: in fact i type the ODE wrongly, but the method should be similar
 
Last edited:
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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