I.V.P. $y'=max\{1,y\}, y(0)=1$

308 Views Asked by Bumbble Comm At 27 Mar 2026 - 7:19

Prove that

have a unique solution then find this solution.

I know I should to prove

i) f(x,y) continues on R.

ii) f(x,y) bounded:

There exist M constant such that:

$|f(x,y)|\leq M$

iii) f(x,y) satisfy lipshtize condition

$|f(x,y_2)-f(x,y_1)|\leq L|y_2-y_1|$

L is lipshtize constant.

But i don't know how can I find $f(x,y)$ here

I say:

$f(x,y)=1, \ if \ y\leq 1$

$f(x,y)=y, \ if \ y>1$

But my teacher say no it's wrong answer.

I know how can i prove the three above conditions but i only want $f(x,y)$

Thanks.