D'Alembert, Lagrange, and Reduction of Order - The Modern Method

The modern technique for reduction of order proceeds as follows.  Given a linear differential equation

$$a_m(x)y^{(m)} + a_{m-1}(x) y^{(m-1)} + \dots + a_1(x) y'+a_0(x)  y = f(x)$$

and a solution $y_1$ to the associated homogenous equation

$$a_m(x) y^{(m)} + a_{m-1}(x) y^{(m-1)} + \dots + a_1(x) y'+a_0(x)  y = 0,$$

suppose that a solution to the non-homogeneous equation has the form $y_2=u y_1$.  Here, $u$ is an unknown function of $x$.  We substitute $y_2$ along with all its derivatives into the first equation.  After carrying out the product rule, every $k$th derivative $(y_2)^{(k)}=(u y_1)^{(k)}$ will have a term of the form $u(y_1^{(k)})$ and these are the only terms containing $u$.  After substituting into the first equation, we collect these $u$ terms to find

$$a_m(x) (u y_1)^{(m)} + a_{m-1}(x)(u y_1)^{(m-1)} + \dots + a_1(x) (u y_1)'+a_0(x)  (u y_1) =$$

$$A(x, y_1, y_1', \dots, y_1^{(m-1)}, u', u'', \dots, u^{(m)}) +$$

$$u\,[a_m(x) (y_1^{m}) + a_{m-1}(x) (y_1^{(m-1)}) + \dots + a_1(x) (y_1')+a_0(x)  y_1] = f(x),$$

where $A$ is a linear differential equation in the unknown function $u$.  Since $y_1$ solves the second equation above, the bracketed expression vanishes.  Hence the first equation above is reduced to

$$A(x, y_1, y_1', \dots, y_1^{(m-1)}, u', u'', \dots, u^{(m)}) =f(x).$$

As there are no $u$ terms in this expression, a change of variables of $w=u'$ gives a linear differential equation for the unknown function $w$ of order $m-1$; namely,

$$A(x, y_1, y_1', \dots, y_1^{(m-1)}, w, w', \dots, w^{(m-1)}) = f(x).$$

This shows we can reduce the order of a linear differential equation if one solution of the associated homogeneous equation is known.  While not often taught, it is also true that if $L_m(y)=0$ is an $m$th order equation with known solutions $y_1, y_2, \dots, y_k,$ we can reduce the order to $m-k$.  Unfortunately, the obvious idea of using all known solutions to successively reduce $L_{m}(y)=0$ won't work since typically none of $y_2,\dots, y_k$ solve $L_{m-1}(y)=0$.

The correct process is described by Ince in [9, p. 121].  Suppose that $k$ solutions $y_1, \dots, y_k$ are known to $L_m(y)=0$.  Use $v_1=y_1$ to reduce $L_m(y)=0$ to $L_{m-1}(y)=0$.  As we mentioned, $y_2$ isn't generally a solution to $L_{m-1}(y)=0$.  However, $v_2=\left(\frac{y_2}{v_1}\right)'=\left(\frac{y_2}{y_1}\right)'$ is a solution, and it can be used to reduce $L_{m-1}(y)=0$ to $L_{m-2}(y)=0$.  Then $v_3=\left(\frac{1}{v_2}\left(\frac{y_3}{y_1}\right)'\right)'$  is a solution to $L_{m-2}(y)=0$, and it can be used to reduce $L_{m-2}(y)=0$ to $L_{m-3}(y)=0$.  Then $v_4=\left(\frac{1}{v_3}\left(\frac{1}{v_2}\left(\frac{y_4}{y_1}\right)'\right)'\right)'$ is a solution to $L_{m-3}(y)=0$, and it can be used to  reduce that equation.  The process continues with as many solutions as were given at the start.