# Thread: First order nonlinear system

1. ## First order nonlinear system

Consider the initial value problem:

y' = 1 + y^2 ; with y = 0 when x = 0.

First part of the question was to get y2(x) by successive approximation. I was able to do that.

I need help on this part:
Let R = [-1, 1] x [-1, 1]. Find the smallest M such that abs(f(x, y)) <= M on R. Find an interval I = (-c, c) such that the graph of every approximation function Yn over I will lie in R.

I'm really stuck on this problem, I don't know where to start.

2. Originally Posted by hashshashin715
Consider the initial value problem:

y' = 1 + y^2 ; with y = 0 when x = 0.

First part of the question was to get y2(x) by successive approximation. I was able to do that.

I need help on this part:
Let R = [-1, 1] x [-1, 1]. Find the smallest M such that abs(f(x, y)) <= M on R. Find an interval I = (-c, c) such that the graph of every approximation function Yn over I will lie in R.

I'm really stuck on this problem, I don't know where to start.
The Idea here is we want to find the maximum value of the derivative on the square because it will give us an upper bound on how fast the function can grow!

Here is an example lets us use the same region R and initial condition, but look at this problem

$y'=x^2+y$ we identify that $f(x,y)=x^2+y$

So on R the maximum will occur at the points $(-1,1);(1,1)$ these give

$|f(x,y)|\le |f(1,1)|=3$ so the maximum slope of this function is $m=3$ since we were given the intial point $y(0)=0$ we can bound this by a linear function

$\displaystyle y' = 3 \implies y = 3x+c \iff y=3x$

So the biggest interval we know for sure that the solution $y$ stays between $\displaystyle -1 < y < 1 \iff -1 < 3x < 1 \iff \frac{-1}{3} < x < \frac{1}{3}$