# ODE question - Integrating factor

1. Feb 3, 2010

1. The problem statement, all variables and given/known data
Find all solutions of the equation:

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

2. Relevant equations
Integrating factor I = exp($$\int$$p(x)dx)

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

3. 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 $$\int$$1/(t.logt)dt

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

dv = (1/t)dt
v = logt

$$\int$$u.dv = uv - $$\int$$v.du

I end up with:

$$\int$$1/(t.logt)dt = 1 + $$\int$$1/(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

2. Feb 3, 2010

### vela

Staff Emeritus
Try the substitution u=log t.

3. Feb 3, 2010

### vela

Staff Emeritus
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.