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.