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

Author(s):
Sarah Cummings (Wittenberg University) and Adam E. Parker (Wittenberg University)

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.

Sarah Cummings (Wittenberg University) and Adam E. Parker (Wittenberg University) , "D'Alembert, Lagrange, and Reduction of Order - The Modern Method," Convergence (September 2015)