Build a fourth-degree series from a differential equation

Denote y=f(x). Both copies give f(0)=1 and the relation $y^{\prime}=2xy-1$. The Teacher completes $f^{\prime}(0)=2(0)(1)-1=-1$. Evaluate derivatives only after differentiating the relation; substituting x=0 into it first would discard the x-dependence needed for later derivatives.

The Student prints the simplified relation; the Teacher fills in the product-rule line:

$\begin{aligned}&y^{\prime\prime}\\ & =\frac d{dx}(2xy-1)\\ & =2x\frac d{dx}(y)+2y\frac d{dx}(x)\\ &{}-\frac d{dx}(1)\\ & =2xy^{\prime}+2y.\end{aligned}$

At zero, $f^{\prime\prime}(0)=2(0)f^{\prime}(0)+2f(0)=2$.

The full Teacher intermediate lines and both source simplifications give

$y^{\prime\prime\prime}=\frac d{dx}(2xy^{\prime}+2y)=2xy^{\prime\prime}+2y^{\prime}+2y^{\prime}=2xy^{\prime\prime}+4y^{\prime},$

$f^{\prime\prime\prime}(0)=4f^{\prime}(0)=-4.$

Differentiating once more,

$y^{(4)}=\frac d{dx}(2xy^{\prime\prime}+4y^{\prime})=2xy^{\prime\prime\prime}+2y^{\prime\prime}+4y^{\prime\prime}=2xy^{\prime\prime\prime}+6y^{\prime\prime},$

$f^{(4)}(0)=6f^{\prime\prime}(0)=12.$

These establish both requested values.

The Teacher completes the full coefficient substitution:

$\begin{aligned}&y\\ & =1+(-1)x+\frac2{2!}x^2\\ &{}+\frac{-4}{3!}x^3+\frac{12}{4!}x^4+\cdots\\ & =1-x+x^2-\frac23x^3+\frac12x^4+\cdots.\end{aligned}$

At x=0.1 the retained polynomial is

$\begin{aligned}&1-0.1+(0.1)^2\\ &{}-\frac23(0.1)^3+\frac12(0.1)^4\\ & =0.909383333\ldots\approx\boxed{0.90938}.\end{aligned}$

This matches the printed five-decimal approximation. Rounding the polynomial is distinct from a separate bound on the omitted terms.