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.