consider:
t*dx/dt = x(1 + log(x/t)), x/t > 0
substitute u= x/t and show that the diff eq reduces to:
t(du/dt) = u*log(u)
please help
t*dx/dt = x(1 + log(x/t)), x/t > 0
substitute u= x/t and show that the diff eq reduces to:
t(du/dt) = u*log(u)
please help
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t dx/dt = x [ 1 + log(x/t)] ---------(1)
let u = x/t
=> x = ut
dx/dt = u + t du/dt
substitute in eqn (1)
t(u + t du/dt) = ut (1 + log(u))
ut + t^2 du/dt = ut _+ ut log(u)
t^2 du/dt = ut log(u)
divide by t
t du/dt = u log(u)
let u = x/t
=> x = ut
dx/dt = u + t du/dt
substitute in eqn (1)
t(u + t du/dt) = ut (1 + log(u))
ut + t^2 du/dt = ut _+ ut log(u)
t^2 du/dt = ut log(u)
divide by t
t du/dt = u log(u)