Solving IVP using Variable Change u = ln(y)

[dm59]

Use the variable change u = ln(y) to solve the IVP dy/dt = -y ln(y), y(1) = 2?

We haven't covered this in class yet so I do not know where to even start.

 

[gabbagabbahey]

Use the variable change u = ln(y) to solve the IVP dy/dt = -y ln(y), y(1) = 2?

We haven't covered this in class yet so I do not know where to even start.

Well, you're given u(t)=ln(y(t)), so why not calculate y(t) in terms of u(t) and then differentiate to get \frac{dy}{dt} in terms of u(t) and \frac{du}{dt} and then substitute these into your differential equation to get a DE in terms of u instead of y. What do you get when you do this?