Solve Initial Value Problem: dy/dt = ty(4-y)/3 with y(0) = a | Calculus Homework

[bodensee9]

Homework Statement

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.

 

[HallsofIvy]

Homework Statement

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

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

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)[/quote ]
y/(4-y) looks better- many people would read y/4-y as (y/4)- y.
Of course, e^((2/3)t^2+ 4C)= (e^(4C))(e^((2/3)t^2)
and since C could be any number, anyway, call that e^(4C), C".
y/(4-y)= C' e^((2/3)t^2)

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.

By "initial value" you mean y(0)= a? Okay, replace t by 0, y by a and solve for C' (or C).

 

[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.

 

[bodensee9]

Got it. Many thanks!