Is there a unique solution to the given ODE?

[TheForumLord]

Homework Statement

Given This ODE:

y' = (y-2) (x^2+y)^5
y(0)=5

A. Show that this problem has one solution that is defined in an open segment that contains 0.

B. Let y(x) be a solution for this problem. Prove that y(x)>2 for every x in I and conclude that y'(x)>0 in I.
Hint: You can use the solution of the problem: y'=(y-2)(x^2+y)^5 , y(x0)=2


Help is needed !

TNX!

Homework Equations

The Attempt at a Solution

 

[HallsofIvy]

This is of the form y'= f(x,y). Your f, $(y-2)(x^2+ y)^5$, is differentiable in both variables in any region that does not include (0,0) so you can use the "existence and uniqueess" theorem.

Further, since y(0)= 5, $y'(0)= (5-2)(0^2+ 5)^5> 0$ and $y'(x)= 0$ only where y= 2 or $y= -x^2$. The latter is impossible so any max or min must be at y= 2. Since the initial value is 5, the derivative is positive there, and can become negative only at y= 2, the function is always larger than 2.

 

[TheForumLord]

Hey there HallsofIvy,
There are some things I didn't understand in your answer:
The initial value is 5 indeed. and from y=5 the function goes up. But how can we know what happened before y=5? Maybe there was a point that was less then y=2? There can be an inflection point in y=2, and then there are values less than y=2...

How can we solve it?


TNX in advance!