MHB How do we deduce that there is such a point?

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evinda
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Hello! (Wave)

Let $k(x)$ and $g(x)$ be continuous functions such that $k(x)>0, 1 \leq g(x) \leq 10$. Show that the problem $ y'(x)=g(x)-k(x)y^2(x), y(0)=0(\star) $ has a global solution.

We firtsly show that if the solution exists then for all $x>0$ it holds $0< y(x) \leq 10 x$.

Indeed , since $y(0)=0$, so $y'(0)=g(0)>0$ and there is a $x_0>0$ such that $y(x)>0$ at $(0,x_0)$.How do we deduce that there is a $x_0>0$ such that $y(x)>0$ at $(0,x_0)$? (Thinking)
 
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evinda said:
Hello! (Wave)

Let $k(x)$ and $g(x)$ be continuous functions such that $k(x)>0, 1 \leq g(x) \leq 10$. Show that the problem $ y'(x)=g(x)-k(x)y^2(x), y(0)=0(\star) $ has a global solution.

We firtsly show that if the solution exists then for all $x>0$ it holds $0< y(x) \leq 10 x$.

Indeed , since $y(0)=0$, so $y'(0)=g(0)>0$ and there is a $x_0>0$ such that $y(x)>0$ at $(0,x_0)$.

Did you mean on $(0,x_0)$? As in, on the interval? I'm going to assume you meant this in the rest of my reply.

How do we deduce that there is a $x_0>0$ such that $y(x)>0$ at $(0,x_0)$? (Thinking)

You would need to prove that $y(x)$ is continuous on at least $(0,x_0)$, if not $[0,x_0]$. You've got $y'(0)>0$, so $y(x)$ must "start out" by increasing.
 
Ackbach said:
Did you mean on $(0,x_0)$? As in, on the interval? I'm going to assume you meant this in the rest of my reply.

Yes, I meant on the interval. (Nod)

Ackbach said:
You would need to prove that $y(x)$ is continuous on at least $(0,x_0)$, if not $[0,x_0]$. You've got $y'(0)>0$, so $y(x)$ must "start out" by increasing.

Why would it suffice to show that $y(x)$ is continuous on at least $(0,x_0)$? Could you explain it further to me? (Thinking)
 
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