D'Alembert, Lagrange, and Reduction of Order - Back to the Classroom

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:

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

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 $\frac{1}{\sqrt{x}}$.

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.