Differential Equations Verifying Implicit Solution

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The discussion revolves around verifying that the equation ln((2X-1)/(X-1))=t is an implicit solution to the differential equation dX/dt=(X-1)(1-2X). The user successfully differentiated the logarithmic equation, leading to the expression dX/dt = -1/((2X-1)(X-1)). They sought guidance on the next steps after deriving this result. Another participant suggested breaking down the differentiation process for clarity and encouraged implicit differentiation to solve for dX/dt. Ultimately, the user confirmed they solved the problem after receiving assistance.
Lancelot59
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I'm given:

1. \frac{dX}{dt}=(X-1)(1-2X)
2. ln(\frac{2X-1}{X-1})=t

and asked to verify that it is an implicit solution to the first order DE given.

I successfully derived the second equation there to get:

\frac{dX}{dt}=\frac{-1}{(2X-1)(X-1)}

So now what? I tried several things and got nowhere.
 
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Lancelot59 said:
I'm given:

1. \frac{dX}{dt}=(X-1)(1-2X)
2. ln(\frac{2X-1}{X-1})=t

and asked to verify that it is an implicit solution to the first order DE given.

I successfully derived the second equation there to get:

\frac{dX}{dt}=\frac{-1}{(2X-1)(X-1)}

So now what? I tried several things and got nowhere.

Show us the work you did when you differentiated (not derived) the second equation.

Note that
ln(\frac{2X-1}{X-1})= ln(2X - 1) - ln(X - 1)

so that should make differentiation a little easier.
 
Mark44 said:
Show us the work you did when you differentiated (not derived) the second equation.

Note that
ln(\frac{2X-1}{X-1})= ln(2X - 1) - ln(X - 1)

so that should make differentiation a little easier.

I did break it up initially:
\frac{dx}{dt}=\frac{d}{dt}( ln(2x-1) - ln(x-1))
\frac{dx}{dt}=\frac{1}{2x-1}(2) - \frac{1}{x-1}(1)=\frac{2}{2x-1} - \frac{1}{x-1}
Then subtracting the fractions:
\frac{dx}{dt}= \frac{2(x-1) - (2x-1)}{(2x-1)(x-1)}=\frac{-1}{(2x-1)(x-1)}
 
Lancelot59 said:
I did break it up initially:
\frac{dx}{dt}=\frac{d}{dt}( ln(2x-1) - ln(x-1))
\frac{dx}{dt}=\frac{1}{2x-1}(2) - \frac{1}{x-1}(1)=\frac{2}{2x-1} - \frac{1}{x-1}
Then subtracting the fractions:
\frac{dx}{dt}= \frac{2(x-1) - (2x-1)}{(2x-1)(x-1)}=\frac{-1}{(2x-1)(x-1)}

Starting from your second equation, you have
t= ln(2x-1) - ln(x-1)
\Rightarrow \frac{dt}{dt}=\frac{d}{dt}( ln(2x-1) - ln(x-1))

You have to differentiate the right side implicitly, and then solve algebraically for dx/dt. Can you take it from here?
 
I solved it! Thanks for the tip.
 
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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