D'Alembert, Lagrange, and Reduction of Order - The Classroom

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. 

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).

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

$$y''+6y'+9y=0.$$

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.