# Initial value problem

1. Jul 18, 2007

### bodensee9

1. The problem statement, all variables and given/known data
Hello, I am wondering if someone can look at the following:

suppose i'm given an initial value problem: dy/dt = ty(4-y)/3 with y(0) = a.

So, if I separable the variables, i would get:

dy/4(4-y) = t/3dt
and I would get by partial fractions
(1/4)*lny-(1/4)ln(4-y) = t^2/6 + C
(1/4)*ln(y/4-y) = t^2/6+C
or (y/4-y) = e^((2/3)t^2 +4C)

so can anyone tell me how I would determine how the value of the solution would vary depending on the initial value a? in other words, how could i get this into a nice form y = ....

Thanks very much.

2. Jul 18, 2007

### HallsofIvy

Staff Emeritus
I presume that is a typo: dy/(y(4-y))= (t/3)dt

Last edited: Jul 19, 2007
3. Jul 18, 2007

### Dick

You can write the solution a little more nicely as y/(4-y)=D*exp((2/3)t^2) (where D=exp(4C) - but it's still just a constant). So your initial value problem is a/(4-a)=D. Solve for a. Clear the fractions, move all the a's to one side etc. You can solve for y(t) in the same way.

4. Jul 18, 2007

### bodensee9

Got it. Many thanks!