# Euler's Method - Global Error

## Homework Statement

Use Euler's method with $$h = 1/2$$ to estimate $$y(1)$$ for the IVP:

$$y(0)=1$$ $$y'(t)=t^2-y(t)$$

Assuming that $$|y(t)| \le 1$$ for $$0 \le t \le 1$$ determine the value of n needed to ensure that $$|E_n| \le 10^{-2}$$

## Homework Equations

$$|E_n| \le \frac{T}{L}(e^{L(t_n-t_0)-1})$$

## The Attempt at a Solution

The first part is easy enough:

$$y_1=y_0+f(t_0,y_0)h=1+f(0,1)(1/2)=1/2$$
$$y_2=y_1+f(x_1,y_1)h=1/2-1/8=3/8$$
$$\Rightarrow y(1)=3/8$$

I'm having trouble with the second part. Could somebody help me out?

Is this correct for L:

$$|f(t,u)-f(t,v)|=|t^2-u-t^2+v|=|u-v|$$

Lipschitz with $$L=1$$

Please, Math Gods, I beg thee...

HallsofIvy
Homework Helper
Is this correct for L:

$$|f(t,u)-f(t,v)|=|t^2-u-t^2+v|=|u-v|$$

Lipschitz with $$L=1$$
Yes, that's true.

Thanks, it's T that I'm having trouble with... how would I find the upper bound for |y''(t)| ?

Is this right?

$$y''(t)=2t-t^2+y(t)$$

So, $$T=(1/2)*(1/2)*y''(1)=(1/4)(2(1)-1^2-3/8)=0.15625$$

and

$$E_n\le|0.15625(e-1)|\Rightarrow E_n\le 0.26848$$

Then set $$E_n=0.01$$

$$\Rightarrow 0.01 = h(0.53696)$$
$$h=0.018623361$$
$$\frac{1-0}{n}=0.018623361$$
$$\Rightarrow n ≈ 54$$

Anyone? Anyone?