I am trying to solve the following ODE: $x^3y''' -3x^2y'' +(6-x^2)xy' -(6-x^2)y=0$ if $y_1=x$. The professor said that the solution could be found by reduction of order. If it's second order the solution this problem is quite easy, but I don't really know how to solve it if it's 3rd order. Any help will be appreciated.

  • 3
    $\begingroup$ Let $y(x)=x z(x)$ and replace. It will be very simple. $\endgroup$ Commented Dec 31, 2021 at 9:54

2 Answers 2


Searching for the other homogeneous solutions by assuming it is the product of some function times another known solution is called variation of parameters. Formally we are plugging in the guess $y = v\cdot y_1 = x\cdot v$ into the equation $D[y] = 0$

$$\begin{cases}y = xv \\ y' = v+xv' \\ y'' = 2v' + xv'' \\ y'' = 3v'' + xv'''\end{cases} \implies x\cdot D[v] + (3x^3v''-6x^2v'+(6-x^2)xv) = 0$$

$$\implies (x^4v'''-3x^3v''+(6-x^2)x^2v'-(6-x^2)xv)+(3x^3v''-6x^2v'+(6-x^2)xv)$$

$$= x^4(v'''-v') = 0$$

In other words $v$ is the three homogeneous solutions to the above differential equation

$$v = C_1 e^x + C_2 e^{-x} + C_3$$

Your professor was justified in calling this a reduction of order because $v'''-v'=0$ is secretly a second order differential equation, with the constant solution already represented by the first homogeneous solution $y_1 = x$. This gives us our final solutions

$$y = C_1 x + C_2 x e^{x} + C_3xe^{-x}$$

  • $\begingroup$ Thanks. I see the point now. Really appreciate your help. $\endgroup$
    – ralfanino
    Commented Dec 31, 2021 at 11:11

It will be easier if you use the formula for the second solution: $y_2=v(x)x$, where x is your first solution.


\begin{equation} \begin{array} fy_2=vx\\ y_2'=v'x+v\\ y_2''=v''x+2v'\\ y_2'''=v'''x+3v'' \end{array} \end{equation}

Insert for every respective term of y in the original ode and get:

$x^3(v'''x+3v'')-3x^2(v''x+2v')+(6-x^2)x(v'x+v)-(6-x^2)vx=0 $

which gives



Sum up each related term and cancel out $-6x^2v'+6x^2v'$ and $3x^3v''-3x^3v''$ and obtain


Divide by $x^4$ and get the simplified form:


Use reduction of order by letting $u''=v'''$ and get:


This has the simple solution $\lambda=\pm\sqrt{-4\cdot-1} \longrightarrow u(x)=e^{2x}+e^{-2x}$. Replace u(x) back to v by integrating it once and get:

$v(x)=\int u(x)dx=\int e^{2x}+e^{-2x}dx=\frac{1}{2}e^{2x}-\frac{1}{2}e^{-2x}+C$

At last, go back to the form of $y_2=v(x)x\rightarrow y_2(x)=\frac{x}{2}e^{2x}-\frac{x}{2}e^{-2x}+Cx$

Nicely put:

\begin{equation} y_2(x)=\frac{x}{2}\bigg(e^{2x}-e^{-2x}\bigg)+Cx \end{equation}

  • 1
    $\begingroup$ I really appreciate your help. I tried the method suggested by Ninad and got the same result as yours. $\endgroup$
    – ralfanino
    Commented Dec 31, 2021 at 11:22

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