Series Solutions — Question 2

Question 2 – Solution

Strategy. Compare the even and odd coefficient chains with an integrated form of the equation.

Step 1: Separate the two chains. Coefficient matching gives $(n+2)(n+1)a_{n+2}+2(n+1)a_n=0$, so $$\boxed{a_{n+2}=-\frac 2{n+2}a_n\quad(n\ge 0).}$$ Thus $u=1-x^2+x^4/2-x^6/6+\cdots$ is even, while $v=x-2x^3/3+4x^5/15-8x^7/105+\cdots$ is odd.

Step 2: Integrate a conserved quantity. Since $(y'+2xy)'=0$, $y'+2xy=C$. At $0$, $C=y'(0)$. Multiplying by $e^{x^2}$ and integrating gives $$y(x)=e^{-x^2}\left[y(0)+y'(0)\int_0^x e^{t^2}\,dt\right].$$ Consequently $\boxed{u=e^{-x^2},\quad v=e^{-x^2}\int_0^x e^{t^2}\,dt}$.

Step 3: Justify convergence and equality. The even coefficients are $(-1)^k/k!$, and the odd ones are $(-2)^k/(2k+1)!!$, where $(2k+1)!!=1\cdot 3\cdots(2k+1)$. Successive nonzero term ratios are $x^2/(k+1)$ and $2x^2/(2k+3)$, respectively; both tend to zero. Thus the series define solutions for all real $x$. The integral formulas solve the same equation with the same initial data, so uniqueness identifies the constructions.

Step 4: Test independence at every point. Differentiating $v$ gives $v'=-2xv+1$, while $u'=-2xu$. Hence $$\boxed{W(u,v)=uv'-u'v=u=e^{-x^2}>0.}$$ At any finite point, the two-by-two system for the coefficients of a combination with zero value and derivative has nonzero determinant. Both coefficients must vanish. The plotted pair remains independent even where one curve crosses zero.