= Solution
Insert the exact solution and expand about $t=t_{n+2}$. Since $f(y)=y'$ and $f'(y)f(y)=y''$, the left side is
$$
y-hy'+\frac25h^2y''.
$$
The coefficient of $h^jy^{(j)}$ on the right side is
$$
\frac45\frac{(-1)^j}{j!}
+\frac15\frac{(-2)^j}{j!}
+\frac15\frac{(-2)^{j-1}}{(j-1)!}
$$
for $j\geq1$, with the last term absent for $j=0$. These coefficients agree through $j=3$; at $j=4$ the left side minus the right side is $1/10$. Thus the <local truncation error> is $h^4y^{(4)}/10+O(h^5)$ and the method has order three.
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