Series Solutions — Question 8

Question 8 – Solution

Strategy. Infer the differential operator from low-order data, then validate all orders by a transformation.

Step 1: Recover the three constants. The constant coefficient in the equation for $u$ gives $-2+\beta=0$. The $x$ coefficient for $v$ gives $-6+\alpha+\beta=0$. The $x^2$ coefficient for $u$ gives $6-2\alpha-\beta+\gamma=0$. These successive equations uniquely force $$\boxed{\beta=2,\qquad\alpha=4,\qquad\gamma=4.}$$

Step 2: State a recurrence valid from the start. With $a_{-2}=a_{-1}=0$ for the shifted contribution, $$\boxed{a_{n+2}=-\frac{(4n+2)a_n+4a_{n-2}}{(n+2)(n+1)}\quad(n\ge 0).}$$ The starting coefficients $a_0=y(0)$ and $a_1=y'(0)$ are free. In particular, $a_2=-a_0$ and $a_3=-a_1$; no negative-index coefficient is an additional parameter.

Step 3: Verify the recovered equation globally. Writing $E=e^{-x^2}$, differentiation gives $$y'=E(w'-2xw),\qquad y''=E\big(w''-4xw'+(4x^2-2)w\big).$$ Thus $y''+4xy'+(2+4x^2)y=Ew''$. The equation becomes $w''=0$, yielding $$\boxed{y=e^{-x^2}(C_1+C_2x),\quad u=e^{-x^2},\quad v=xe^{-x^2}.}$$ Their initial data match the measured series, and their Wronskian is $E^2=e^{-2x^2}>0$. Both functions are entire, so both Taylor radii are infinite. The transformation also proves the entire measured series, not just the first displayed terms, are consistent.

Step 4: Test a new exact measurement. The exponential expansion gives $[x^8]u=1/4!=1/24$ and $[x^9]v=1/24$. The recurrence produces the same values. Since the recovered parameters and each solution’s initial data are fixed uniquely, $1/20$ is incompatible; it cannot be accommodated by choosing later coefficients independently.