Explanation of why Lagrange error formula for specific differential equation does not work

81 Views Asked by At

This excerpt comes from the textbook "Fundamentals of Differential Equations and Boundary Value Problems 7th Edition"

"The most transparent is due to Lagrange: if the (n+1)st derivative of f exists and is continuous on an interval containing $x_0$ and x, then $\epsilon_n(x)=\frac{f^{n+1}(\xi)(x-x_0)^{n+1}}{(n+1)!}$ where $\xi$ although unknown is guaranteed to lie between $x_0$ and $x$.

Figure 8.1 on page 422 and equation (6) suggest that one might control the error in the Taylor polynomial approximation by increasing the degree n of the polynomial (i.e., taking more terms), thereby increasing the factor (n + 1)! in the denominator. This possibility is limited, of course, by the number of times f can be differentiated. In Example 2, for instance, the solution did not have a fifth derivative at $x_0$ = 0 ($f^{(5)}(0)$ is “infinite”). Thus, we could not construct $p_5(x)$, nor could we conclude anything about the accuracy of $p_4(x)$ from the Lagrange formula."

Example 2: Initial value problem $$y''=3y'+x^2y; y(0)=10, y'(0)=5$$

The textbook then deduced that $y^{(5)} =3y^{(4)}+6y'+6xy''+x^2y'''$ and $y^{(5)}(0)=495$

So, my confusion arises because the excerpt stated that the solution did not have a fifth derivative but it does in this example. Is this a mistake on the textbooks part or am I misunderstanding?