Solve the differential equation: (x * y' - 1) * ln(x) = 2 * y
it's already been differentiated?
i dont understand what you mean.
nvm...do you know implicit differentiation?
yeah
ok that makes it easier...I'll calc it in a second
substituting \[x := e^t\] seems to make it a lot easier
the derivative of ln(x) is 1/x, but we have to do product rule
sorry man not sure on this one
and gotta get back studying for lin alg final -___-
try www.wolframalpha.com for an answer
oh ok thanks for your help :) @ nowhereman: if i do that i get y'*(e^t)*t=2y
I got \[(\frac{dy}{dt} - 1)t = 2y\]
assuming the function is analytic, that can be solved with power series
would i be able to have \[(dy/dt) - (2/t)y=1\] and then do an integrating factor:\[e^(intergral of (-2/t))\]?
no, I don't see how that would work.
... i dont understand what i should do then
umm and i dont get how you got the dt part
I used the chain rule \[\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}\]
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