I need help with differential equations?

Question

the differential equation is:

u'=u(u-1) and the initial condition is u(0)= u sub 0

Thank you

Answer 1

This is a separable first order equation:

du/dt = u*(u-1)

du/(u*(u-1)) = dt

ln(u-1) - ln(u) =  ln((u-1)/u) = t + C

where C is a constant to be determined from the initial condition u(0) = u0

Plug in t=0 to get:

ln((u0 - 1)/u0) = C

so

ln((u-1)/u) = t + ln((u0 - 1)/u0)

ln((u-1)/u) - ln((u0 - 1)/u0) = t

ln((u-1)*u0/((u0-1)*u)) = t

((u-1)/u)*(u0/(u0-1)) = exp(t)

u - 1 = u*((u0-1)/u0)*exp(t)

u*(1 - ((u0-1)/u0)*exp(t)) = 1

u(t) = 1/(1 - ((u0-1)/u0)*exp(t))

You can check solution to make sure that it both satisfies the initial condition and the original differential equation.  First check to see that u(0) = u0:

u(0) = 1/(1 - ((u0-1)/u0)*exp(0))

u(0) = 1/(1 - ((u0-1)/u0)*1)

u(0) = 1/(1 - 1 +1/u0)

u(0) = u0  check.

Now take the derivative of u(t)

u(t) = 1/(1 - ((u0-1)/u0)*exp(t))

To save time and space in writing, let's define a = ((u0-1)/u0)

u(t) = 1/(1 - a*exp(t))

u'(t) = (a*exp(t))/(1-a*exp(t))^2

Now, at first glance, this doesn't look like the original differential equation, but note that this can be rewritten as:

u'(t) = (a*exp(t))*(u(t))^2

and if we rearrange the solution to solve for a*exp(t), we get:

u(t) = 1/(1 - a*exp(t))

1 - a*exp(t) = 1/u(t)

a*exp(t) = 1 - 1/u(t)

a*exp(t) = (u(t) - 1)/u(t)

so

u'(t) = (a*exp(t))*(u(t))^2 = ((u(t) - 1)/u(t))*u(t)^2

u'(t) = u(t)*(u(t) - 1)

which is the original differential equation, so this is the correct solution.