Just want to understant(Diff EQ)

1. Sep 8, 2005

karen03grae

Ok guys, this is not a difficult problem at all (ch 1)...sorry to disappoint you. But after 3 lectures I still can't answer parts of these questions -and I have the solution's manual :( Here they go:

Does the initial value problem y' =3*y^2/3 , y(0)=10^-7

have an unique solution in a neighborhood of x=0?.

Okay, I applied the Existence and Uniqueness theorem and took the partial derivative of y' with respect to y and got 2*y^-1/3. Now since my starting point is (0,10^-7), my y' is continuous so it passes the test. And my i.v.p. does in fact have a unique solution at that starting point.

Okay, now don't you find it weird that my functions aren't depending on x?

And the solutions manual goes on to say both functions y' and partial of y' w.r.t. y are defined on this rectangle R={(x,y): -1<x<1, (1/2)10^-7 <y<(2)10^-7} What?!? Where did those numbers come from? Why can't x be equal to or greater than 1? My y' and partial of y' w.r.t. y don't even depend on x so why is x bounded?

Any help will be greatly appreciated, as always...Karen

2. Sep 8, 2005

Fermat

If you integrate y', you'll see that y does depend on x.

3. Sep 8, 2005

karen03grae

yeah, i thought about that...but even so why can't x be -1 or 1...or bigger than 1? y(x)= 3*y^2/3*x. And how did they get that interval.

4. Sep 8, 2005

HallsofIvy

Staff Emeritus
Your problem is not differential equations, it's calculus!

The integral of y'= 3y3/2 is not y= 3y3/2x!

y is not a constant and you can't treat it like one. If you're already working with the "existance and uniqueness" theorem, then you should already have learned about "separable equations". y'= 3y3/2 is the same as
$$\frac{dy}{dx}= 3y^{\frac{3}{2}$$
which is equivalent to (in "differential" form)
$$y^{-\frac{3}{2}}dy= 3dx$$.
Integrating that, the left side wrt y and the right side wrt x, gives
$$2y^{-\frac{1}{2}}= 3x+ C$$
or y= 4/(3x+C)2. Notice that y cannot be negative! That's because there is a problem with the differential equation at y=0.

You are correct that since (3y3/2)'= (9/2)y1/2 is continuous for y not 0, the given initial value problem will have a unique solution- although possibly only on a very short interval!

5. Sep 8, 2005

karen03grae

Okay I asked my teacher and he told that the chosen interval was probably just random.