How can I solve this nonlinear ODE homework on Apollo reentry?

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SUMMARY

The discussion focuses on solving a nonlinear ordinary differential equation (ODE) related to Apollo reentry, specifically addressing the challenge of handling a single-variable equation where acceleration is defined as the second derivative (a = d²s/dt²) and velocity as the first derivative (V = ds/dt). The participant struggles with integrating factors and substitution methods, ultimately finding a workaround by using the relationship dv/ds = (dv/dt)*(dt/ds) = (dv/dt)*1/V. This approach leads to a separable differential equation for dv/ds, allowing for the solution of velocity in terms of position, which can then be substituted back to find acceleration.

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Students in advanced mathematics or physics courses, particularly those studying differential equations, as well as engineers and scientists involved in aerospace dynamics and reentry analysis.

Mr.Waka
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Homework Statement


http://www.math.udel.edu/~moulton/Apollo%20EC.pdf
This is the full problem that I am working on for my ODE class.

Homework Equations


I would figure acceleration equals the second derivative so a=d^2s/dt^2
and V=ds/dt like the hint says.

The Attempt at a Solution


I looked over the examples of Nonlinear ODEs but they all have 2 variables instead of just one like this one. I was looking for an integrating factor but none of that works out I think. I tried to integrate it by parts twice but that lead to a jumbled mess. I tried to subsitute s in terms of V but that didnt work to well either.

For me the only trouble that I am having is the fact that its a nonlinear problem with really only one variable because V is in terms of S

And looking at a,b and c all of them don't deal with the s variable. I'm not too sure where in the math that it gets canceled out. In fact each step seems to reduce the variables needed.

Im just not sure how to tackle the problem. The math shouldn't be bad but I can't find a place to start.
 
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A common trick you can use is that dv/ds = (dv/dt)*(dt/ds) = (dv/dt)*1/v = d^2 s/dt^2 * 1/v. Since d^2 s/dt^2 is given in terms of v and s, you get a differential equation for dv/ds which you can solve easily by separating. After solving for v you plug that in the original equation to get an equation for a in terms of s. Then you can find the maximum of a by finding what a is when a'(s) = 0. When I did this, the answer I got for part a) was off from what they got by a factor of 1/2, however I got the same thing they did for part b)...
 

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