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.

We now share questions about using the technique of reduction of order that we hope are helpful even for those well versed in the method. 

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In practice, this method is often used only to reduce a second order linear differential equation as follows.

Question 1.

Given that \(y_1=x^{-1}\) is a solution of the linear differential equation

\[2x^2 y''+xy'-3y =0,\]

find the second solution.

Solution 1. Assume the second solution can be written in the form \(y_2=u x^{-1}.\)  Substituting \(y_2'=u' x^{-1}-u x^{-2}\) and \(y_2'' = u''x^{-1}-2u'x^{-2}+2ux^{-3}\) gives

\[2x^2 (u''x^{-1}-2u'x^{-2}+2ux^{-3})+x(u' x^{-1}-u x^{-2})-3ux^{-1} =0.\]

Since \(y=x^{-1}\) solves the equation \[2x^2 y''+xy'-3y =0,\] we have

\[2x^2 (x^{-1})''+x (x^{-1})'-3x^{-1} =2x^2 (2x^{-3})+x (-x^{-2})-3x^{-1} = 0.\]

Substituting this into \[2x^2 (u''x^{-1}-2u'x^{-2}+2ux^{-3})+x(u' x^{-1}-u x^{-2})-3ux^{-1} =0\] gives the differential equation

\[2u''x-4u'+u'=0.\]

Letting \(w=u'\) we have

\[2w' x - 3w =0,\] which is a separable first order linear equation and gives  \(w=x^{3/2}\).  So \[u=\int w\, = \frac{2}{5} x^{5/2}.\] Thus \[y_2=u y_1 = \frac{2}{5}x^{5/2}x^{-1} = \frac{2}{5}x^{3/2},\] which is our second solution (up to a constant).

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An unfortunate consequence of students only ever reducing second order equations \[a_2(x) y''+a_1(x) y'+a_0(x) y=0\]

is that often they simply memorize the formula [15, p. 135]

\[u=\int\frac{e^{-\int \frac{a_1(x)}{a_2(x)}dx}}{y_1^2(x)}dx,\]

which masks the full power of the method.  We can avoid memorization by asking questions that naturally generalize the second order process given in a textbook.  For example, we can start with a differential equation of order \(>2\).

Question 2. Given that \(y_1=e^x\) is a solution to \(y'''+3y''-4y = 0,\) reduce the order to a second order equation.

Solution 2. Since \(y_1=e^x\) is a solution, we assume another solution is of the form \(y_2=uy_1\).  Then, substituting \(y_2'=u'y_1+uy_1'\), \(y_2''=u''y_1 + 2 u'y_1'+uy_1'',\) etc. into the differential equation gives

\[u'''+6u''+9u'=0\]

or the second order equation

\[y''+6y'+9y=0.\]

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A slightly more difficult problem is to reduce the order when multiple solutions are provided.

Question 3. Given that \(y_1=e^x\) and \(y_2=xe^{-2x}\) are solutions to

\[y'''+3y''-4y = 0,\]

find the third solution.

Solution 3. As in Question 2, we start with \(y_1=e^x\) and suppose the third solution has the form \(y_3=uy_1.\)  Substitution into the differential equation shows \(u\) must satisfy

\[u'''+6u''+9u'=0\]

or if \(w=u'\)

\[w''+6w'+9w=0.\]

We now restart the process, knowing that \(w_1=\left(\frac{y_2}{y_1}\right)'\) solves \[w''+6w'+9w=0.\] This means \(\frac{y_2}{y_1}\) solves \[u'''+6u''+9u'=0\] and could play the role of \(u\) in \(y_3=uy_1\).  Unfortunately, doing so gives \(y_3=u y_1 = \left(\frac{y_2}{y_1}\right)y_1= y_2\) and returns a known solution.

Rather, proceed as in Question 1, and suppose the second solution to \[w''+6w'+9w=0\] is of the form \[w_2=vw_1=v\left(\frac{y_2}{y_1}\right)'=v(xe^{-3x})'\] for some unknown function \(v\).  Substitution into \[w''+6w'+9w=0\] gives

\[(1-3x)v''-6v'=0\]

which has a solution

\[v=\frac{c_1}{1-3x}+c_2\]

and so

\[w_2 =vw_1= \left(\frac{c_1}{1-3x}+c_2 \right)\left(\frac{y_2}{y_1}\right)'=\left(\frac{c_1}{1-3x}+c_2 \right)(xe^{-3x})'\]

\[= c_1e^{-3x}+(1-3x)c_2 e^{-3x}.\]

Thus \(u=c_1e^{-3x}+c_2 xe^{-3x}\) which gives \[y_3 = e^x(c_1e^{-3x}+c_2 xe^{-3x})=c_1e^{-2x}+c_2xe^{-2x}.\] Since we were given that \(y_2=xe^{-2x}\) is a solution, we can disregard it and find that  \(y_3=e^{-2x}.\)

Now that we have thoroughly reviewed the modern technique and its uses in the classroom, we discuss its origins.

Our modern method for reduction of order is due to the French mathematician and physicist Jean le Rond d'Alembert (1717-1783) in 1766 [3, p. 381].  Figure 1 shows d'Alembert published exactly our method.

Your problem on integrating the equation \(Py+\frac{Q dy}{dx}+\frac{Rd^2y}{dx^2} \dots + \frac{M d^my}{dx^m}=X\)  when one has \(m-1\) values of \(x\) [sic] in the equation \(Py+\frac{Q dy}{dx}+\frac{Rd^2y}{dx^2} \dots + \frac{M d^my}{dx^m}=0\) seems so beautiful to me that I've looked for a solution myself, which follows. Let \(y=V\xi\), \(V\) being undetermined, and \(\xi\) one of the values of \(y\) that satisfies the equation \(Py+\frac{Q dy}{dx}+ \dots \& c.=0\) and so this value be substituted in the equation \(Py+\frac{Q dy}{dx}+ \& c. \dots=X\).   The transformed will be composed of, 1) one part \(V(P\xi+\frac{Q d\xi}{dx}\dots + \frac{M d^m\xi}{dx^m})\)… where \(X\) does not exist, so it will evidently \(=0\), because \(P\xi +\frac{Q d\xi}{dx}\dots + \frac{M d^m\xi}{dx^m}=0\)...

Figure 1. D'Alembert explained what remains the modern method for reducing the order of a linear differential equation in 1766 [3, p. 381]. D'Alembert's French is followed by the authors' translation into English.

What is interesting, and what provides motivation for the rest of this paper is the title of his article:  “Extrait de différentes lettres de M. d'Alembert à M. de la Grange écrites pendant les années 1764 & 1765" (“Excerpt from different letters between Mr. d'Alembert and Mr. de LaGrange written during the years 1764 & 1765”).  It appears that this technique comes from a conversation with Joseph-Louis Lagrange (1736-1813).  To learn more, we examine the letters shared between these mathematicians.

Lagrange and d'Alembert started their correspondence in 1759.  At this time, d'Alembert was 41 and one of the preeminent living mathematicians.  Lagrange was 22 and already establishing himself as a first class mathematician in his own right.  A year earlier, Lagrange and his students started the Turin Academy of Sciences and published the first issue of the Mémoires of their new society.  This journal went by many names over the years, but today is known as Miscellanea Taurinensia.  On September 27, 1759, d'Alembert wrote to Lagrange from Paris and began by complimenting him on the first issue (see Figure 2). This letter started a correspondence that lasted for 24 years. 

Monsieur, I've received the first volume of your Memoires that you have done the honor of sending me with much gratitude and read with much satisfaction. Your essay on sound is full of the most scholarly and ingenious research.

Figure 2. The beginning of years of correspondence between d'Alembert and Lagrange [12, p. 1].  In each of Figures 2 through 6, D'Alembert's or Lagrange's French is followed by the authors' translation into English.  (Source: Excerpts from the correspondence of d'Alembert and Lagrange reproduced in Figures 2 through 6 are from a copy of Lagrange's Oeuvres [12] held by Bibliothèque nationale de France, département Littérature et art, V-15599 (BIS).)

The first mention of reduction of order occurred in a letter from Lagrange to d'Alembert on January 26, 1765 (see Figure 3).  In this letter, Lagrange told d'Alembert he had a technique of reducing the order of a linear differential equation and moreover he planned to publish his method in the upcoming third volume of Miscellanea Taurinensia.  Notice Lagrange did not provide details of his method here. 

\(X\) being some function of \(x\), and \(y=A+Bx\); and this solution in and of itself is only a specific case for an integration method of which I calculate the full value of \(y\) in this equation of \(m\)th degree: \[Py+Q\frac{dy}{dx}+R\frac{dy^2}{dx^2}+\dots = X\] \((P, Q, \dots, X\) being arbitrary functions of \(x)\) when assuming that I know \(m\), or least \(m-1\), specific values of \(y\) in the equation \[Py+Q\frac{dy}{dx}+R\frac{dy^2}{dx^2}+\dots = 0.\] This will be the subject of a memoire that I will insert in the third volume of our Mélanges.

Figure 3. Lagrange alluded to a method to reduce the order, but gave few details [12, pp. 30-31].

Lagrange continued by complimenting d'Alembert's methods on the famous three body problem, then asked if d'Alembert would be willing to publish them in the upcoming issue as well (see Figure 4). 

Your method of integration for the three bodies problem is extremely simple and convenient; I will send you mine, which is completely different, as soon as you wish; allow me to assure you in advance, it will not displease. Our society is preparing to publish a new volume: would you like to do it the honor of putting your name on it? This would certainly have a great effect, and it would speed up its acceptance in the community. Send us some of your papers; I will put them in order and have them printed with all due care.

Figure 4. Lagrange asked d'Alembert to publish his studies on the three body problem in Miscellanea Taurinensia [12, p. 31].

D'Alembert responded in March of 1765.  He described (in more detail than Lagrange shared) his own method for reduction of order.  It's clear this is our modern process (see Figure 5).

I've provided, in the first edition of my Traité de Dynamique, a method for integrating the equation \[ddy+Mydz^2+Pdz^2 = 0\] which allows me very simply to integrate the equation \[Py+Q \frac{dy}{dx}+R\frac{d^2y}{dx^2} +\dots = X\] when one has \(m-1\) values of \(y\) in \(x\) in the case of \(X=0\). So \(y=\nu z\), \(z\) being one of the values and \(\nu\) an undetermined variable; I have \[Pz+Q\frac{dz}{dx} + R\frac{d^2 z}{dx^2} + \dots = 0\] and by substitution, it gives me a degree \(m\) equation which will only have \(\nu\) as an unknown  and no term at all with \(V\); so, having \(\frac{dv}{dz}=q\) I will have an equation of \(m-1\) degree where I will have \(m-2\) values of \(q\) in the case of \(X=0\), since \(\nu\) or \(\frac{y}{z}\) has \(m-2\) known values (hyp.) and as a result, \(\frac{dv}{dz}\) or \(q\); therefore by continuing in this way, I will arrive at an equation of the form \[dr +Z r dz+\zeta dz = 0\] which is evidently integrable.

 Figure 5.  D'Alembert responded with his own method for reduction of order [12, pp. 33-34].

He also addressed Lagrange's request for a paper (see Figure 6), and mentioned conflicts with the Academies of Paris and Berlin which made him hesitant to publish with Miscellanea Taurinensia.  This may not be a surprise considering historian of science Thomas Hankins’ statement [8, p. 28] that “In pure contentiousness d'Alembert came in a close second to the Bernoullis; whatever he lacked in viciousness he made up in the dogged determination with which he pursued his pet theories and claims of priority."  The story of these conflicts is interesting on its own, though it is not highly relevant to Lagrange’s role in developing the reduction of order technique.  Please see the Appendix for details on these controversies. 

In regard to your proposal to me my dear and illustrious friend, to insert a piece about my method in the Memoirs, it’s an honor which I appreciate very much and which I very much desire to be able to respond, but since I want to avoid any difficulties with the Academy, to whom I no longer send any memoires for the reasons of which you are already aware, and the same with the Academy of Berlin, where for a long time I have no longer sent anything either. Here’s what I could to do: Write you a long letter in which I would summarily treat different matters and in which (which is more important and dear to me) I would have the opportunity to do you, without it seeming to flatter, the justice you deserve. You could give this writing the title “Extrait de différentes lettres de M. d'Alembert à M. de la Grange.” This would be like a kind of analysis of the main things that I must treat in the fourth volume of my Opuscules. If it suits you, tell me in what time to have it ready. You can count on my word, provided I have the time before me, because I neither want to nor am I able to hurry. My health does not allow me to complete extensive work on the same material.

Figure 6.  D'Alembert proposed a compromise concerning contributing to Miscellanea Taurinensia [12, pp. 34-35].

D'Alembert did not completely reject Lagrange's request to contribute to Miscellanea Taurinensia, but rather proposed a compromise.  He agreed to send some notes to Lagrange, which Lagrange could organize and publish under the title “Extrait de différentes lettres de M. d'Alembert à M. de la Grange.”  In March 1765, Lagrange accepted the offer, referring to d'Alembert's reduction of order technique as being “très-belle” and noted that his method is different [12, p. 37].  In June, d'Alembert asked for a concrete deadline [12, p. 40] and in July, Lagrange said he needed it by the end of the year [12, p. 42].  On September 6, he reminded d'Alembert again, and on September 28, d'Alembert promised the material by the end of December [12, p. 45].  D'Alembert was true to his word and sent the notes with plenty of time to spare ... on December 28.

It appears that the relationship between mathematicians, editors, and deadlines hasn't changed much in 250 years.

As promised, Lagrange published his own method in the next issue of Miscellanea Taurinensia [11, p. 179].  His “De différens problèmes de calcul intégral” (“On different problems on integral calculus”) immediately preceded d'Alembert's paper in the journal (see Figure 7).  In order to make his method more understandable, we'll adopt the notation proposed by historian S. S. Demidov in [4, p. 370]. 

On the integration of the equation \[Ly + M \frac{dy}{dt}+N\frac{d^2y}{dt^2}+P\frac{d^3y}{dt^3}+\&c. = T   \hspace{1 cm}(A)\] in which \(L,M,N \text{ etc. } T\) are functions of \(t\). 1. I multiply this equation by \(\xi dt\), \(\xi\) being an undetermined variable, I take the integral, I have \[\int L \xi y dt + \int M \xi \frac{dy}{dt} dt+ \int N \xi \frac{d^2y}{dt^2} dt +  \int P \xi \frac{d^3y}{dt^3}dt  + \&c. = \int T \xi dt ;\] I change the expressions \(\int M \xi \frac{dy}{dt} dt, \quad \int N \xi \frac{d^2y}{dt^2} dt ,\quad \int P \xi \frac{d^3y}{dt^3} dt\)  to their equivalents  \(M \xi y - \int \frac{d M \xi}{dt} y\,dt,\)    \(N \xi \frac{dy}{dt} - \frac{d N \xi}{dt} y + \int \frac{d^2 N \xi}{dt^2} y\,dt,\)    \(P \xi \frac{d^2y}{dt^2}-\frac{d P \xi}{dt} \frac{dy}{dt}+ \frac{d^2 P \xi}{dt^2} y - \int \frac{d^3 P \xi}{dt^3} y\,dt\,\,\&c.\)  And I have, by ordering the terms by relation to \(y,\) ....

Figure 7. Lagrange began his article on his method for reduction of order [11, p. 179].  Lagrange's French is followed by the authors' translation into English.

Suppose we are given a linear differential equation of the form

\[L_m(y) = a_m(x) y^{(m)} +a_{m-1}(x) y^{(m-1)} + \dots + a_0(x) y = X(x)\]

and we know that \(y_1\) is a solution to the associated homogenous equation \(L_m(y)=0\).

Lagrange multiplied both sides of this differential equation by an undefined function \(z\) and integrated, giving

\[\int z\, [a_m(x) y^{(m)} +a_{m-1}(x) y^{(m-1)} + \dots + a_0(x) y ] dx = \int  z X(x) dx.\] He executed \(k\) integrations by parts on each \(z a_{k}(x) y^{(k)},\) which left  \((z a_{k}(x))^{(k)} y\) under the integral.  For example when \(k=2\), \[\int z a_{2}(x)   y'' dx = z a_{2}(x)  y' -\int (z a_{2}(x))' y' dx\]

\[= z a_{2}(x)  y'-(z a_{2}(x))' y+\int (z a_{2}(x))'' y \, dx.\] Doing this for each term on the left side and collecting terms under the integral gave

\[\int z L_n(y) dx  = A(x,y,y', \dots, y^{m-1}, z, z', \dots, z^{(m-1)}) + \int y L_m^*(z) dx\] \[= \int z X(x) dx.\] The expression \(A(x,y,y', \dots, y^{m-1}, z, z', \dots, z^{(m-1)})\) is called the bilinear concomitant and is a bilinear function in \(y\) and \(z\) and their derivatives. (In general a concomitant is “a phenomenon that naturally accompanies or follows something.”) Finally, the adjoint \(L_m^*(z)\) of the differential operator \(L_n(y)\) is given by

\[L_m^*(z) =(-1)^m (za_m(x))^{(m)}+(-1)^{(m-1)}(z a_{m-1}(x))^{(m-1)} +\dots - (z a_{1}(x) )' + z a_0(x).\]

If \(L_m=L_m^*\), then the operator is self-adjoint [16, p. 96], in which case any solution \(y_1=z_1\) of  \(L_m(y)\) is also a solution to \(L_m^*(z)\).  It follows that the integral \(\int y L_m^*(z_1)dx =0\) and the equation

\[A(x,y,y', \dots, y^{m-1}, z, z', \dots, z^{(m-1)}) + \int y L_m^*(z) dx = \int z X(x) dx\] would simplify to the \((m-1)\)th order linear equation

\[A(x,y,y', \dots, y^{m-1}, z_1, z_1', \dots, z_1^{(m-1)}) + 0 = \int z_1 X(x) dx,\] as desired. 

Further reduction can be accomplished when additional solutions are known.  If \(L\) is self-adjoint and \(z_1, z_2, \dots, z_k\) are \(k\) solutions to \(L_m^*(z)=0\) (since \(k\) solutions \(y_1, y_2, \dots, y_k\) are known to \(L_m(y)=0\)), we obtain a system of \(k\) linear differential equations

\(A(x,y,y', \dots, y^{m-1}, z_1, z_1', \dots, z_1^{(m-1)}) + 0\)	\(= \int z_1 X(x)\)
\(A(x,y,y', \dots, y^{m-1}, z_2, z_2', \dots, z_2^{(m-1)}) + 0\)	\(= \int z_2 X(x)\)
	\(\vdots\)
\(A(x,y,y', \dots, y^{m-1}, z_k, z_k', \dots, z_k^{(m-1)}) + 0\)	\(= \int z_k X(x)\)

and \(y^{(m-1)}, y^{(m-2)}, \dots, y^{(m-k+1)}\) can be eliminated to obtain a differential equation of order \((m-k)\) [15, p. 124].

Certainly not every operator is self-adjoint, though those that describe physical processes often are. Finding conditions for an operator to be self-adjoint is easy.  By equating the coefficients in the differential equation \[L_m(y) = a_m(x) y^{(m)} +a_{m-1}(x) y^{(m-1)} + \dots + a_0(x) y = X(x)\] and with those of the adjoint \[L_m^*(z) =(-1)^m (za_m(x))^{(m)}+(-1)^{(m-1)}(z a_{m-1}(x))^{(m-1)} +\dots - (z a_{1}(x) )' + z a_0(x),\] the following theorem arises:

Theorem [16, p. 98]. The second-order linear differential equation

\[a_2(x)y''+a_1(x)y'+a_0(x)y = 0\] is self-adjoint if \(a_1(x)=a_2'(x)\) and the third-order linear differential equation

\[a_3(x)y'''+a_2(x)y''+a_1(x)y'+a_0(x)y=0\]

is self-adjoint if \(a_2(x)=\frac{3}{2}a_3'(x)\) and \(a_0(x)=\frac{1}{2}(a_1(x)-\frac{1}{3}a_2'(x))'\).

When the differential equation is not self-adjoint, Lagrange repeated the process on \(L_m^{*}(z)=0\), or equivalently

\[(-1)^m (za_m(x))^{(m)}+(-1)^{(m-1)}(z a_{m-1}(x))^{(m-1)} +\dots - (z a_{1}(x) )' +z a_0(x)  =0.\] He multiplied both sides by an unknown function \(w(x)\) and integrated.  Executing multiple integrations by parts gives

\[\int w L_m^{*}(z) dx = B(x,z,z',\dots,z^{(m-1)}, w, w', \dots, w^{(m-1)}) + \int z L_m^{**}(w) dx\]

\[= \int z \cdot 0 \, dx = c,\] where \(L_m^{**}\) is the adjoint of the adjoint of the original operator.  Lagrange proved that \(L_m^{**}=L_m\).  Hence a solution, \(w_1\), to the differential equation \(L_m^{**}(w) =0\) is known; namely \(w_1=y_1\).  Plugging this into the preceding equations shows that

\[B(x,z,z',\dots,z^{(m-1)}, w_1, w_1', \dots, w_1^{(m-1)}) =c,\] which is a differential equation of order \(m-1\).  If we integrate this equation to find a solution, \(z_1\), then we have that \(L_m^{*}(z_1)=0\).  Finally, by the equation \[A(x,y,y', \dots, y^{m-1}, z, z', \dots, z^{(m-1)}) + \int y L_m^*(z) dx = \int z X(x) dx,\] we have \[A(x,y, y', \dots, y^{(m-1)}, z_1,z_1',\dots,z_1^{(m-1)}) = \int z_1 X(x) dx,\]

which is also a differential equation of order \(m-1\).

In general, the authors agree with Demidov, who stated, “D'Alembert's method was simpler and more convenient than Lagrange's; no wonder it is widely used today” [4, p. 372].  However, when \(L\) is self-adjoint, the problems are easy and appropriate for undergraduates to solve using Lagrange's method.

Question 4.  Given that \(y_1=x^{-1}\) is a solution to the third order self-adjoint linear differential equation

\[2x^3 y'''+9x^2 y''+6xy'=0,\]

find a second order differential equation.

Solution 4. If \(L_3(y) = 2x^3 y'''+9x^2 y''+6xy'\), then the above process gives \[\int z\,[2x^3 y'''+9x^2 y''+6xy'] dx =\]

\[\underbrace{(2x^3 zy''+(9x^2 z-(2x^3 z)')y'+((2x^3 z)''-(9x^2z)'+6xz)y}_{A(x,y,y',y'',z,z',z'')}+\int y L_3^*(z) dx\] \[= \int z \cdot 0 = k.\]

Since \(L_3^*  = L_3\) (checking this makes a good problem as well),  \(z= x^{-1}\) makes the integral \(\int y L_3^*(z) dx\) vanish.  We are left with bilinear concomitant and second order equation:

\[A(x,y,y',y'',x^{-1},(x^{-1})',(x^{-1})'') = 2x^2 y'' +5xy'+y = 0.\]

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If Lagrange's method has a benefit, it is when multiple solutions are known.  This is particularly useful in the self-adjoint case. 

Question 5.  Given that \(y_1=x^{-1}\) and \(y_2=c\) are solutions to the third order self-adjoint linear differential equation

\[2x^3 y'''+9x^2 y''+6xy'=0,\] find all solutions.

Solution 5.  If \(L_3(y) = 2x^3 y'''+9x^2 y''+6xy'\), then the above process with \(y_1=x^{-1}\) gives \[\int z\,[2x^3 y'''+9x^2 y''+6xy'] dx =\]

\[\underbrace{(2x^3 zy''+(9x^2 z-(2x^3 z)')y'+((2x^3 z)''-(9x^2z)'+6xz)y}_{A(x,y,y',y'',z,z',z'')}+\int y L_3^*(z) dx\] \[= \int z \cdot 0 = k.\]

Since \(L_3^*  = L_3\),  \(z= x^{-1}\) makes the integral \(\int y L_3^*(z) dx\) vanish.  We are left with bilinear concomitant and second order equation

\[A(x,y,y',y'',x^{-1},(x^{-1})',(x^{-1})'') =2x^2 y'' +5xy'+y = 0.\]

Similarly, the above process with \(y_1=c\) leaves a bilinear concomitant and second order equation

\[A(x,y,y',y'',c,c',c'') =2x^2 y'' +3xy'= 0.\]

Combining  the two preceding equations to eliminate \(y'',\) we find

\[2xy'+y =0,\]

which gives the third solution \(\frac{1}{\sqrt{x}}\).

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Simply having students execute Lagrange's method for non self-adjoint differential equations can be challenging, though appropriate for enrichment projects or to motivate integral transforms later in the course.

Question 6.  Given that \(y_1=x^{-1}\) is a solution of the second order differential equation \[y''+\frac{1}{2x}y'-\frac{3}{2x^2} y =0,\] find the second solution.

Solution 6.  We take the above equation, multiply both sides by an unknown function \(z(x),\) and integrate by parts multiple times to get that

\[\int z\left[y''+\frac{1}{2x}y'-\frac{3}{2x^2} y\right] dx = \int z \cdot  0 \,dx\] and

\[\underbrace{z y' - z' y +\frac{z y}{2x}}_{A(x,y,y', z, z')} + \int y[\underbrace{z''-\left(\frac{z}{2x}\right)'-\frac{3 z}{2x^2}}_{L_2^*(z)}]dx = c_1.\]

Simplification shows that \[L_n^*(z) = z''-\left(\frac{1}{2x}\right)z' - \frac{1}{x^2} z\] and we apply the above method to the differential equation \(L_2^*(z)=0\).  We multiply both sides by an unknown function \(w\), integrate, and apply integration by parts multiple times to get

\[\int w\left[z''-\left(\frac{1}{2x}\right)z' - \frac{1}{x^2} z\right] dx = \int w \cdot 0 \, dx\] and

\[\underbrace{wz'-zw'-\frac{wz}{2x}}_{B(x,z,z', w, w')} + \int z[\underbrace{w''+\left(\frac{w}{2x}\right)'-\left(\frac{w}{x^2}\right)}_{L_2^{**}(w)}]dx = c_2.\]

Now \(L_2^{**}(w)=0\) is the same differential equation as what we started with.  And so \(w=x^{-1}\) solves it, meaning the integral will evaluate to zero.  Hence we have reduced the problem to solving \[\frac{1}{x} z' - \left(\frac{-1}{x^2}\right) z - \frac{z}{2x^2} =c_2,\] which is a first order linear differential equation with solution \(z_1=  \frac{2 c_2}{5} x^2+ \frac{c_3}{\sqrt{x}}\).  Hence \(L_2^*(z_1) = 0\) and so \(\int y L_2^*(z_1)=0.\)  This means that \(A(x,y,y',z_1,z_1') = c_1\) and

\[y \left(\frac{c_3}{x^{3/2}}-\frac{3 c_2 x}{5}\right)+\left(\frac{c_3}{\sqrt{x}}+\frac{2 c_2 x^2}{5}\right) y' =c_1,\]

which has a solution

\[y(x) = \left(\frac{c_4 \left(5 c_3+2 c_2 x^{5/2}\right)}{x}-\frac{c_1}{c_2 x}\right) = Cx^{-1} + Dx^{\frac{3}{2}},\]

as desired.

Leonhard Euler (1707-1783) (Wikimedia Commons, public domain)

Whenever recovering mathematics from the 1700s, it is always prudent to check the work of Leonhard Euler to make sure he didn't do it earlier or better.  Indeed when looking at Euler's contributions to the study of linear differential equations, one can find the techniques of both d'Alembert and Lagrange.  First, in 1743, Euler published “De integratione aequationum differentialium altiorum graduum” (“On the integration of differential equations of various degrees”) [5, p. 193].  He described how to solve higher order linear differential equations with constant coefficients

\[0=Ay + \frac{B dy}{dx} + \frac{C ddy}{dx^2}+ \frac{Dd^3y}{dx^3} + \frac{E d^4 y}{dx^4} + \&c.\]

Euler assumed the solution was of the form \(e^{zx}\) and substituted into this differential equation, giving an algebraic auxiliary equation

\[0=A+Bz+Cz^2+Dz^3+Ez^4+ \&c.\]

If \((pz-q\)) is a factor of this equation, then \(e^{\frac{qx}{p}}\) is a solution to the differential equation.  Euler was aware of this as far back as 1739 when he and Johann Bernoulli corresponded on these equations [7, p. 448].  His 1743 paper also answers what happens when \((pz-q)\) is a repeated factor.  If \((pz-q)^2\) is a factor of the auxiliary equation, then Euler let \(y=e^{\frac{px}{q}} u\) where \(e^{\frac{px}{q}}\) is the known solution and \(u\) is an unknown function.  He proceeded to show that \(u=\alpha + \beta x\), which gave him the second solution to the equation.  Later in this paper he studied equations with higher order repeated factors as well as those with complex roots.

Figure 8. Euler used d'Alembert's technique in 1743 to deal with repeated roots of the auxiliary equation [5, p. 204].

This seems to turn the tables on the typical relationship between Euler and d'Alembert (see the Appendix).  This time Euler earned priority, but d'Alembert presented the material in greater generality and with more clarity.

Morris Kline, in his Mathematical Thought from Ancient to Modern Times, mentioned that Euler also utilized Lagrange's method, though well after Lagrange published it in 1766.  He wrote [10, p. 487]:

Hence Lagrange discovered the theorem that the adjoint of the adjoint of the original non-homogeneous ordinary differential equation is the original but homogeneous equation.  Euler did essentially the same thing in 1778.  He had seen Lagrange's work, but apparently forgot about it.

It appears Kline was referring to Euler’s paper “Observatio singularis circa aequationes differentiales lineares” (“Special observations about linear differential equations”), a paper written in 1778, but published years later in 1805 [6, p. 42].

Conclusion

Utilizing history in the classroom is an important technique that helps improve learning.  When we are lucky, we discover new mathematics in the process.  When we are very lucky, that mathematics can be used in the classroom, which is what happened in this project. 

Lagrange first brought the problem of reduction of order to d'Alembert's attention, a problem d'Alembert called "... so beautiful to me that I've looked for a solution myself."  Lagrange already had a solution and it didn't take long for d'Alembert to find his own.  D'Alembert often rushed to publication to guarantee priority to the detriment of clarity.  However, on the topic of reduction of order, the tables were turned.  D'Alembert lost priority to Lagrange, but his method is so clear it remains with us today.

Acknowledgments

The authors would like to thank all the people who made suggestions that greatly improved this paper.  In particular the referees and editor made many helpful comments, and Dr. Timothy Wilkerson of Wittenberg helped review several of the French translations.

About the Authors

Sarah Cummings (Wittenberg University) is a 2015 graduate of Wittenberg University in Springfield, Ohio, where she was a Mathematics major and French minor.  She was thrilled to see her two passions, math and the French language, come together in this project unraveling the history behind methods for reduction of order.  Sarah is currently living in Chicago, Illinois, and is a first year graduate student in DePaul University’s Predictive Analytics program.

Adam E. Parker (Wittenberg University) is associate professor of mathematics at Wittenberg University.  He has undergraduate degrees in mathematics and psychology from the University of Michigan and earned his mathematics Ph.D. in 2005 from the University of Texas at Austin under the direction of Dr. Sean Keel.  He teaches a wide range of classes and often tries to incorporate primary sources in his teaching.  This paper grew out of just such an attempt.  In his free time he enjoys sports, cooking, and repairing mechanical watches.

Here, we discuss the controversies d'Alembert alluded to in Figure 6.  The first seemed to stem from a competition with Alexis Clairaut.  Clairaut was elected to the Academy of Sciences in 1731 at the age of 18, and d'Alembert followed in 1741 at the age of 24.  The competition started in 1742 when d'Alembert began to present his Traité de Dynamique (Treatise on Dynamics), which he refered to in Figure 5.  A month later, Clairaut presented his own work on dynamics, Quelques Principles qui Donnent la Solution de Plusieurs Problèmes de Dynamique (Some Principles Which Give the Solution to Many Problems in Dynamics).  D'Alembert must have been intimidated by the title and content in Clairaut's early lectures.  Fearing losing his claim to priority, he immediately took his work to the Secretary of the Academy of Sciences to acknowledge it was complete.  He then rushed to private publishers so that it would appear before the issue of Mèmoires of the Paris Academy which contained Clairaut's work [8, p. 30-31].  Not surprisingly, the work was unready for publication.  According to Amir Alexander in Duel at Dawn, “The result is a work that has been much praised for its insights, but also much criticized for being obscure and unreadable” [1, p. 27].  The competition escalated and became unprofessional, with each making petty, condescending insults to the other, often in press.   

Thomas Hankins, in Jean D’Alembert: Science and the Enlightenment [8], gave an excellent description of the entire feud between d'Alembert, Clairaut, and the Academy in general, complete with a fine bibliography.  He noted [8, p. 40]:

As with most scientific controversies it is difficult to say who was right and who was wrong.  Public opinion was with Clairaut, because he had correctly predicted the comet's return, while d'Alembert merely quibbled over an incomprehensible mathematical theory.  Also d'Alembert's ‘public image’ was at a low point because of the collapse of the Encyclopèdie and the hostility of many journalists who were receiving the favor of the court.

As such, d'Alembert fell out with many of his colleagues at the Academy and publishing there ceased to be an option, even though he remained a member and in fact was elected perpetual secretary from 1772 until his death in 1783 [13]. 

Clairaut died on May 17, 1765, and one month later d'Alembert complained bitterly to Lagrange that Clairaut's pension (of 9000 to 10000 francs) had not been passed to himself quickly enough.  He felt it was his “vested right as the oldest member."  He said he was used to being treated poorly by the academy, but acknowledged, “The academy, who must fear losing me, finally wrote the minister to ask that the pension go to me."  Several months later the pension was indeed passed to him [12, p. 38].

The disagreement with the Berlin Academy is no less petty, though perhaps a bit more clear.  Euler was the mathematical titan of his day, if not of all time.  He was the chief mathematician at the Academy of Berlin from 1741 to 1765.  Inevitably, he and d'Alembert worked on some of the same problems, including the three body problem that interested Clairaut as well.  Indeed, given d'Alembert's propensity to publish prematurely,

... his chief innovations all appeared in much clearer and more accurate form in the works of Euler. …  It was d'Alembert's fate to see most of his ‘inventions' adopted by Euler and given a much more satisfactory treatment than he himself was capable of giving them.  [8, p. 42]

Given d'Alembert's personality, he fought hard to make sure that his original ideas were recognized as such, even forcing Euler to issue an apology stating that he “did not make the slightest pretension to the glory that was due to d'Alembert" [1, p. 28].

D'Alembert couldn't compete with Euler mathematically.  However, he did compete politically.  King Frederick II was the patron of the Berlin Society and he was enamored with the high society of Paris and France.  In fact, he hired Pierre-Louis Moreau de Maupertius away from the Paris Academy to be the President of the Berlin Academy in 1746 [1, p. 29].  Maupertius was an early advocate for d'Alembert in Paris, and his departure certainly caused d'Alembert's relationship with the Academy to deteriorate.  However, his arrival in Berlin created an ally for d'Alembert to Frederick. 

Even before Maupertius left Berlin in 1756, Frederick attempted to hire d'Alembert as the President of the Berlin Society.  This position would naturally have gone to Euler and in fact many of the responsibilities did.  However, none of the “benefits or honors" were bestowed [1, p. 30].  D'Alembert declined Frederick's offer, seemingly because of a genuine desire to stay in the more cosmopolitan Paris from which he rarely traveled.  Euler, however, thought his declination was merely an attempt to negotiate a larger salary from Frederick, at which time Euler essentially banned d'Alembert from the competitions and journals of the Berlin Society.  According to Amir Alexander [1, p. 30]:

This last was a particularly severe blow because, thanks in large part to Euler, the Berlin Histoire has become the leading scientific journal in Europe, and d'Alembert has long preferred it to the Paris Academy's Mémoires de l'Académie Royale des Sciences.  With no place to publish his work, d'Alembert resorted to publishing his mathematical papers in his own series Opuscules Mathématiques.

[1]  A. Alexander. Duel at Dawn: Heroes, Martyrs, and the Rise of Modern Mathematics (dissertation). President and Fellows of Harvard College, Boston, 2010. Also available in revised version, Harvard Univ. Press, 2011.

[2]  W. E. Boyce and R. C. DiPrima. Elementary Differential Equations. Wiley, New York, 10th edition, 2012.

[3]  J. L. R. d'Alembert. Extract de différentes lettres de M. d'Alembert à M. De La Grange écrites pendant les années 1764 & 1765. Misc. Taurinensia, 3:381-396, 1766. Available online via Google Books.

[4]  S. S. Demidov. On the history of the theory of linear differential equations. Arch. Hist. Exact Sci., 28:369-387, 1983.

[5]  L. P. Euler. De integratione aequationum differentialium altiorum graduum. Misc. Berolinensia, 7:193-242, 1743. Available online at MAA Euler Archive as E62.

[6]  L. P. Euler. Observatio singularis circa aequationes differentiales lineares. Nova Acta Academiae Scientarum Imperialis Petropolitinae, 14:52-61, 1805. Available online at MAA Euler Archive as E720.

[7]  J. Fauvel and J. Gray. The History of Mathematics: A Reader. Macmillan Press, London, 1987.

[8]  Thomas L. Hankins. Jean d'Alembert: Science and the Enlightenment. Gordon and Breach Science, New York, 1970.

[9]  E. L. Ince. Ordinary Differential Equations. Dover Publications, New York, 1944.

[10]  M. Kline. Mathematical Thought From Ancient to Modern Times, volume 2. Oxford Univ. Press, 1972.

[11]  J. L. Lagrange. Solution de différens problêmes de calcul intégral. Misc. Taurinensia, 3:179-380, 1766. Available online via Google Books.

[12]  J. L. Lagrange. Oeuvres, volume 13. Gauthier-Villars, Paris, 1882. Our source was a copy held by Bibliothèque nationale de France (BNF), département Littérature et art, V-15599 (BIS), available online via Gallica.

[13]  J.J. O’Connor and E.F. Robertson, Jean Le Rond d'Alembert, MacTutor History of Mathematics Archive. October 1998.

[14]  D. E. Otero. History helps make sense. Math Horizons, pp. 10-20, April 2015.

[15]  D. Zill. A First Course in Differential Equations: The Classic Fifth Edition. Cengage, Pacific Grove CA, 2000.

[16]  D. Zwillinger. Handbook of Differential Equations. Academic Press, Waltham Mass., 3rd edition, 1997, reprinted 2007. Available online in pdf format.