2.1 Theorem:

Let $y_i \,\,\,(1\leq i \leq n)$ be the approximations of a solution y of (2.1) at $x_i$.Let the (exact)solution y be twice continually differentiable on [a,b], $(x_0 = a,\,\, x_0 + nh = b)$. Further let

$$\vert f_y(x,y)\vert\leq L\,\,$$    and $$\,\ \vert y''(x)\vert\leq \vert M\vert, \, \, \, a\leq x \leq b$$

where L and M are positive constants. Then

$$\vert e_i\vert \leq \frac{hM}{2L}(e^{ihL}-1)$$ (2.3)

Proof:
By the mean value we have,

$$y(x_{i+1})=y(x_i)+hy'(x_i)+\frac{h^2}{2!}y''c),\,\,\, x_i \leq c\leq x_{i+1}$$

we also know,

$$y_{i+1}=y_i+hf(x_i,y_i)$$

Substraction, now leads to,

$$e_{i+1}=e_i+h\{f(x_i,y(x_i))-f(x_i,y_i)\}+\frac{h^2}{2!}y''i)$$ (2.4)

Again, by mean value theorem

$$f(x_i,y(x_i))-f(x_i,y_i)=f_y(x_i,d)(y(x_i)-y_i)$$

$$\qquad\qquad\qquad = f_y (x_i, d). e_i$$

where d lies between $y_i$ and $y(x_i).$ Substituting in (2.4)we get

$$\vert e_{i+1}\vert\leq \vert e_i\vert+h \, \vert e_i\vert+\frac{h^2}{2}M$$ (2.5)

Let $g_{i+1}$ be the solution of the difference equation

$$g_{i+1}=(1+hL)gi + \frac{ h^2}{2}M, \,\, g_0 =0..$$ (2.6)

Claim:

$$g_i \geq \vert e_i\vert, \,\,\, i=1,2,...n$$

The claim follows by induction. Also by induction

$$g_i=A(1+hL)^i-A, \,\,\, A=\frac{hM}{2L}$$

$$\qquad\qquad\qquad \qquad \leq A (e^{nhi}-1)$$

$$\qquad\qquad\qquad \qquad \quad\,\,\,\,\,\,=\frac{hM}{2L}(e^{(x_i -a)}-1)$$

The theorem now is proved once we notice $\vert e_i\vert\leq d_i,\,\,\,i=1,2,....n$
Remark: Inequality (2.3) implies that the error is $O(h^2)$. Theorem also given an upper band for the error.