Domain of solution to Cauchy prob.

[Kalidor]

Prove that the solution of the CP

$$y'=-(x+1)y^2+x$$
$$y(-1)=1$$

is globally defined on all of $$\mathbb{R}$$

How would you go about this? I thought about studying the sign of the right member if the equation. But what would I do next?

 

[HallsofIvy]

What are you allowed to use? The fact that both $f(x,y)= -(x+1)y^2+ x$ and f_y(x,y)= -2(x+1)y[/math] are continuous for all (x, y) and the fundamental existence-uniqueness theorem should do it.

 

[Kalidor]

Hi HallsofIvy.
I guess I'm only allowed to use the theorem of local existence and uniqueness and the fact that (f being locally lipschitz and continuous) if for any compact K there exists a constant \(\displaystyle C \) such that
$$|f(x,y)| \leq C(1+|y|) \forall x \in K$$ and $$y \in \mathbb{R}^n$$
then there is a global solution, otherwise this exercise wouldn't be marked as "pretty hard" in my book.