1. ## ODE-Lipschitz&Picard–Lindelöf theorem

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!

2. For part a, as you mentioned,
we can use the...

Picard–Lindelöf theorem

...which says that, as $\displaystyle f(x,y)=(y-2)(x^2+y)^5$ is continuous in x and Lipschitz continuous in y, there is a unique solution to the initial value problem, which is defined in a neighbourhood $\displaystyle J$ of 0.

Now for part b, note that $\displaystyle y_0=2$ is a solution of the equation. If $\displaystyle y$ was to meet $\displaystyle y_0$ somewhere over $\displaystyle J$, the uniqueness of the initial value problem $\displaystyle \{z'=f(x,z), \ z(0)=2, x\in \mathbb{R}\}$ would be violated. Since $\displaystyle y(0)>2$, we must have $\displaystyle y>2$ throughout $\displaystyle J$. Now the ode gives us a constant sign for $\displaystyle y'$ over $\displaystyle J$.

3. Thanks a lot