Reduction of Order for Linear Equations

Definition Reduction of order starts from the normalized homogeneous equation

on an open interval $I$ on which $p$ and $q$ are continuous. An equation $P(x)\,y'' + Q(x)\,y' + R(x)\,y = 0$ is first divided by $P(x)$ on an interval where $P(x) \ne 0$, which gives $p = Q/P$ and $q = R/P$. A known solution is a solution $y_1$ on $I$ that is not the zero function. It may come from inspection, from a trial such as $x^r$ or $e^{rx}$, or from an earlier method, and it is confirmed by substitution. A known solution can vanish at points of $I$, but only at isolated points. At a zero $x_0$ the slope $y_1'(x_0)$ cannot also be zero, since uniqueness would then make $y_1$ the zero function, so near $x_0$ the graph crosses the axis only once.

Definition Let $y_1$ be a known solution of $y'' + p\,y' + q\,y = 0$ on an open interval $J$ on which $p$ and $q$ are continuous and $y_1$ has no zeros. Reduction of order looks for solutions of the form $y = v(x)\,y_1(x)$ with an unknown twice differentiable function $v$. The product rule gives

After substitution, the terms that contain $v$ without a derivative add up to $v\,(y_1'' + p\,y_1' + q\,y_1)$, which is zero because $y_1$ is a solution. The equation becomes

Since $v$ itself no longer appears, the function $w = v'$ satisfies the first-order linear equation

and this drop from order two to order one gives the method its name. With the integrating factor $y_1^2e^{\int p\,dx}$, its solutions are

for a constant $C$.

Definition On an open interval $J$ on which $p$ and $q$ are continuous and the known solution $y_1$ has no zeros, a second solution is

where one fixed antiderivative of $p$ is used in the exponent and any antiderivative of the quotient is used outside. With definite integrals from a point $x_0$ of $J$ the formula reads

with $E(s) = e^{-\int_{x_0}^{s} p(r)\,dr}$. It is the case $C = 1$ of the previous card. Only the normalized coefficient $p$ enters, and $q$ is used only through the fact that $y_1$ is a solution.

Definition Let $y_2 = v\,y_1$ with $v' = e^{-\int p\,dx}/y_1^2$ on $J$, as in the formula. Then

which is never zero. So $y_1$ and $y_2$ are linearly independent on $J$ and form a fundamental pair there, and every solution on $J$ is $C_1y_1 + C_2y_2$. This agrees with Abel’s identity. Adding a constant $c$ to $v$ replaces $y_2$ by $y_2 + c\,y_1$, which is another valid second solution with the same Wronskian. The constant $C$ in $w = C\,e^{-\int p\,dx}/y_1^2$ must be nonzero, since $C = 0$ makes $v$ constant and returns a multiple of $y_1$.

Definition The substitution divides by $y_1$, and the formula has $y_1^2$ in a denominator, so reduction of order is carried out on an open interval $J \subseteq I$ with $y_1(x) \ne 0$ for every $x$ in $J$. Because the zeros of a known solution are isolated, removing them from $I$ leaves open intervals of this kind. A zero of $y_1$ restricts the method, not the equation. Since $p$ and $q$ are continuous on all of $I$, the existence and uniqueness theorem for linear equations shows that every solution found on $J$ is the restriction of exactly one solution on $I$, and the simplified formula often shows that extension directly. A point where $P(x) = 0$, or where $p$ or $q$ fails to be continuous, is a singular point of the equation itself, and there the theorem gives no such guarantee.