Series Solutions — Question 4

Question 4 – Solution

Strategy. Expand every term in the shifted variable and keep the forcing coefficients in the same index convention.

Step 1: Match the forcing coefficient. Derivatives with respect to $x$ and $h$ agree. The coefficient of $h^n$ in $hy'$ is $na_n$ for $n\ge 0$; it is zero when $n=0$. Thus $$\boxed{a_{n+2}=\frac{1/n!-na_n}{(n+2)(n+1)}\quad(n\ge 0),\qquad a_0=0,\ a_1=1.}$$

Step 2: Compute the initial segment. Successive applications give $a_2=1/2$, $a_3=0$, $a_4=-1/24$, $a_5=1/120$, $a_6=1/144$. Therefore $$\boxed{P_6(h)=h+\frac{h^2}{2}-\frac{h^4}{24}+\frac{h^5}{120}+\frac{h^6}{144}.}$$ The cubic coefficient vanishes because $1/1!-a_1=0$, not because the solution has a fixed parity.

Step 3: Interpret the residual correctly. Direct differentiation cancels the coefficients through $h^4$. At degree five, $hP_6'$ contributes $1/24$ while $e^h$ contributes $1/120$; hence $$\boxed{\mathcal R(h)=\frac{h^5}{30}+O(h^6).}$$ The omitted term $a_7h^7$ would contribute $42a_7h^5$ to $y''$. Indeed, the recurrence gives $a_7=-1/1260$, cancelling that residual term. Two differentiations lower the degree of an omitted term by two.

Step 4: Recover and validate the exact solution. Writing $v=y'$ gives $v'+hv=e^h$. An integrating factor yields $$\boxed{y(2+h)=\int_0^h e^{-s^2/2} \left(1+\int_0^s e^{t+t^2/2}\,dt\right)ds.}$$ At $h=0$ the outer integral is zero and its integrand is $1$, verifying the initial values. Differentiating gives the first-order equation for $v$, hence the original equation. Its leading coefficient is $1$ and all coefficients and forcing are entire, so the stated theorem applies to disks of every radius. The Taylor series is therefore entire.