who can give the exact answer to this problem? first to answer correctly gets "best answer" no matter what

Question

Use the change of variable z = ln(y) to solve the nonlinear IVP

(dy/dt) + y = yln(y) , y(0) = 1

y(t) = ???

yes helmut very nice. you get best answer.

Answer 1

(dy/dt) + y = yln(y)
let z = ln(y)
y = e^z
dy = e^zdz
(dy/dt) + e^z = ze^z
(dy/dt) = e^z(z - 1)
dy = e^z(z - 1)dt
e^zdz = e^z(z - 1)dt
dz = (z - 1)dt
dz/(z - 1) = dt
ln(z - 1) = t + C
z - 1 = Ce^t
ln(y) = 1 + Ce^t
0 = 1 + C
C = - 1
ln(y) = 1 - e^t
y = e^(1 - e^t)

Answer 2

dy/dt + y = yln(y)

The solution is y(t)=(e)^(((e)^t C+1))

With the initial condition y(0) = 1,

y(t)=(e)^(((e)^((t+I Pi))+1))