Given a general solution, find the EDO

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SUMMARY

The discussion revolves around finding the ordinary differential equation (EDO) from the general solution given by the formula y = (2*c*e^(2x)) / (1+c*e^(2x)). The user attempted to isolate c*e^(2x) using implicit differentiation and derived an EDO: y = [y'/(4-y'-2*y)]*(2-y). However, upon using Mathematica's DSolve function, the user received a different general solution. The consensus is that there is no error in the user's approach; rather, DSolve provides an equivalent representation of the solution.

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1. The problem statement

Given the general solution:
y==(2*c*e^(2x)) / (1+c*e^(2x))
find the EDO.

2. The attempt at a solution

Im tried isolate c*e^(2x), using implicit differentiation:

y + y*c*e^(2x)==2*c*e^(2x)
y'+y'*c*e^2x + 2*y*c*e^(2x)==2*2*c*e^(2x)
y'==c*e^(2x) * (4-y'-2y)
e^(2x) * c = y'/(4-y'-2*y)

Replacing in the general solution given in the problem statement i have this EDO:

y==[y'/(4-y'-2*y)]*(2-y)

But, in mathematica software when i use DSolve over this EDO to find the general solution, it gives a different one.

What's wrong?

Thanks!
 
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scarebyte said:
1. The problem statement

Given the general solution:
y==(2*c*e^(2x)) / (1+c*e^(2x))
find the EDO.

2. The attempt at a solution

Im tried isolate c*e^(2x), using implicit differentiation:

y + y*c*e^(2x)==2*c*e^(2x)
y'+y'*c*e^2x + 2*y*c*e^(2x)==2*2*c*e^(2x)
y'==c*e^(2x) * (4-y'-2y)
e^(2x) * c = y'/(4-y'-2*y)

Replacing in the general solution given in the problem statement i have this EDO:

y==[y'/(4-y'-2*y)]*(2-y)

But, in mathematica software when i use DSolve over this EDO to find the general solution, it gives a different one.

What's wrong?

Thanks!

There is nothing wrong. I suspect that DSolve is just giving a different, equivalent way of writing the solution.
 
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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