ODE question - Integrating factor

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The discussion centers on solving the differential equation y' = (2y)/(t.logt) = 1/t for t > 0 using the integrating factor method. The user identifies p(t) as -2/(t.logt) and attempts to find the integrating factor I through integration by parts. They encounter a paradoxical result of 0 = 1, indicating a mistake in handling the integration. A suggestion is made to use the substitution u = log t to resolve the issue, emphasizing the importance of including the constant of integration in indefinite integrals. The conversation highlights common pitfalls in applying integration techniques to differential equations.
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Homework Statement


Find all solutions of the equation:

y' = (2y)/(t.logt) = 1/t, t > 0


Homework Equations


Integrating factor I = exp(\intp(x)dx)

where y' + p(x)y = q(x)


The Attempt at a Solution


Hi everyone, here's what I've done so far:

Let p(t) = -2/(t.logt)

I = exp(\int((-2)/(t.logt))dt)

Factoring out the -2, consider \int1/(t.logt)dt

Use integration by parts:
u = 1/logt
du = -(logt)^-2.(1/t).dt

dv = (1/t)dt
v = logt

\intu.dv = uv - \intv.du

I end up with:

\int1/(t.logt)dt = 1 + \int1/(t.logt)dt

which gives me 0 = 1, which is clearly wrong.

But I can't see where I'm going wrong! I've done it five times now and I keep getting the same answer. Can anyone see where I'm going wrong or suggest another way of solving the problem?

Thanks for any help
 
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Try the substitution u=log t.
 
You didn't do anything wrong when you integrated by parts. If you were evaluating definite integrals, the first term would drop out since you'd get 1-1=0. With the indefinite integrals, the paradoxical result arises because you've neglected the constant of integration.
 
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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