= Solution
The origin is an <ordinary point>, since the coefficients of the <second-order linear differential equation> are <polynomials>. Seek a <power series> $y=\sum_{n\geq0}a_nx^n$. Matching the coefficients of $x^0$ and $x^1$ gives $a_2=a_3=0$. Matching the coefficient of $x^{j+2}$ gives
$$
(j+4)(j+3)a_{j+4}+(4j+1)a_j=0,
\qquad
\boxed{a_{j+4}=-\frac{4j+1}{(j+4)(j+3)}a_j\quad(j\geq0).}
$$
Thus the four residue classes of indices modulo four evolve separately. The <initial conditions> for the first solution give $a_0=1$, $a_1=0$, and hence
$$
y_1(x)=\sum_{m=0}^{\infty}(-1)^m
\left(\prod_{j=0}^{m-1}\frac{16j+1}{(4j+4)(4j+3)}\right)x^{4m}
=1-\frac{x^4}{12}+\frac{17x^8}{672}-\frac{17x^{12}}{2688}+\cdots.
$$
The second set of <initial conditions> gives $a_0=0$, $a_1=1$, and
$$
y_2(x)=\sum_{m=0}^{\infty}(-1)^m
\left(\prod_{j=0}^{m-1}\frac{16j+5}{(4j+5)(4j+4)}\right)x^{4m+1}
=x-\frac{x^5}{4}+\frac{7x^9}{96}-\frac{259x^{13}}{14976}+\cdots.
$$
An empty product equals one. In either <power series>, the ratio of successive nonzero terms tends to zero for every fixed $x$: the coefficient ratio is of order $1/m$, while the power increases by four. The <ratio test> therefore proves convergence for every $x$, and termwise differentiation verifies the <differential equation>. These normalized solutions are <linearly independent>.
For their <Wronskian> $W=y_1y_2'-y_1'y_2$, differentiating and substituting the <differential equation> gives
$$
W'=y_1y_2''-y_1''y_2=-4x^3W,\qquad W(0)=1.
$$
Thus the <Abel identity> yields
$$
\boxed{W(x)=e^{-x^4}.}
$$
On the interval containing zero where $y_1$ has no zero,
$$
\left(\frac{y_2}{y_1}\right)'=\frac{W}{y_1^2}=\frac{e^{-x^4}}{y_1(x)^2}.
$$
Since $y_2(0)/y_1(0)=0$, integration proves the requested <reduction of order> formula:
$$
\boxed{y_2(x)=y_1(x)\int_0^x\frac{e^{-\xi^4}}{y_1(\xi)^2}\,d\xi.}
$$
This is a local formula around the origin, as requested. The <power series> define both solutions globally; an ordinary integral through a zero of $y_1$ is not intended. <Reduction of order across a zero of the known solution> instead uses continuation of the actual solution.
Back to article page
