Check Differential Equation Solution

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The discussion revolves around verifying the solution to the differential equation dX/dt=(2-x)(1-x) with the proposed solution ln((2-x)/(1-x))=t. The user initially struggles with substituting the derivative into the original equation but eventually identifies the correct common denominator as (2-x)(1-x). After correcting their approach, they successfully derive that dX/dt=(2-x)(1-x) and confirm that this satisfies the original differential equation. The final conclusion is that the proposed solution is indeed valid for the differential equation.
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Homework Statement


Verify Differential Equation Solution
dX/dt=(2-x)(1-x)

Homework Equations


The solution is ln ((2-x)/(1-x))=t


The Attempt at a Solution


The derivative of the solution is -1/(2-x) dX/dt - (-1)/(1-x) dX/dt=1
But I plug this derivative in and I get stuck. How am I supposed to plug the solution and the derivative of the solution into the original differential equation? Can you help me set this up? I'm going to need a little help along the way...
 
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-1/(2-x) dX/dt - (-1)/(1-x) dX/dt=1

rewrite it as

\left( \frac{-1}{2-x}+ \frac{1}{1-x} \right) \frac{dX}{dt}=1

and bring the two terms in the bracket to the same common denominator.
 
Thanks for the reply. I did that and my common denominator is (2-x).

With that, I solve for dX/dt and I get dX/dt=1/(2-x).

I plug the dX/dt into the original equation which looks like 1/(2-x)=(2-x)(1-x)

the (2-x) cancels from both sides and I am left with 1=1-x and then x=0.

I went wrong somewhere...
 
I apologize...I found my mistake. The common denominator is
Code:
X[SUP]2[/SUP]-3x+2
which of course factors to (2-x)(1-x). I then multiply the numerator and denominator by the common denominator and put everything together. I then multiply both sides by (2-x)(1-x) to solve for dX/dt which leaves me with dX/dt=(2-x)(1-x). Plugging this into the original DE leaves me with (2-x)(1-x)=(2-x)(1-x) which proves that the solution is a true solution for this DE. Thanks for all of your help...
 
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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