# When solving a third-order ODE using reduction of order, what happens when the v’ term drops out

#### Given question

Given $$y=x$$ is a solution of $$6x^3y’’’ - 24x^2y’’ + 48xy’-48y = 0$$. Find the general solution if $$x>0$$.

#### My work

Since we are given that $$y = x$$ is a solution, my first approach would be to use reduction of order method. So, I would say that

\begin{align} y_2 &= v*y_1 \newline y_2’ &= v + v’y_1 \newline y_2’’ &= v’ + v’ + y_1v’’ = 2v’ + y_1v’’ \newline y_2’’’&= 3v’’ + y_1v’’’ \end{align}

When I plug this into the original equation I get

$$6x^4v’’’ - 6x^3v’’ = 0$$

Normally, when I use this method the v term will drop making my homogenous equation a function of $$\frac{dv}{dx}$$’s but here my v term and v’ term dropped out. From some googling I found the following

“If you’ve done all of your work correctly this should always happen. Sometimes, as in the repeated roots case, the first derivative term will also drop out” but I’m not quite sure what this means for my work….

I’m used to making the substitution $$w=v’$$. Could someone point me in the right direction?

• I saw this one, but I focused on the second comment. In the first comment, they just made the substitution $v = y’’$. Which makes it first-order. After solving the first order equation, would I just need to take the integral twice? Commented May 28 at 15:12
• Yes, at the end, you will have to integrate the result for $v$ twice, in order to recover $y$. Note that your differential equation turns out to be an instance of Euler's equation (see en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation), whose usual method of solving starts from the monomial ansatz $y(x) = Ax^n$. Commented May 28 at 15:43

## 2 Answers

The process is essentially the same: Just set $$w = v''$$, which transforms the new equation (after clearing common factors) to $$x w' - w = 0 ,$$ which is first order and separable and has solution $$w(x) = C x$$. (You can just as well view this substitution as setting $$u = v'$$ as usual, noticing that the equation is $$u$$ doesn't have a $$0$$th order term, and so setting $$w = u'$$.) Now, integrate twice to recover $$y_2(x)$$, which introduces two more constants of integration.

Alternatively, that every term has the form $$x^m y^{(m)}$$ suggests the ansatz $$y(x) = x^a$$, since after substituting each term is some multiple of $$x^a$$.

In our case, substituting gives $$6 (a - 1) (a - 2) (a - 4) x^a = 0$$, giving the solution $$y(x) = A x + B x^2 + C x^4 .$$ Notice, by the way, that the original differential equation is $$-x$$ times the Wronskian determinant of the functions $$y(x), x, x^2, x^4$$.

The differential operator

$$d=x^k \partial_{k_k}$$

is reproducing powers with factors.

So simply solve

    d = Function[y ,6 x^3 y'''[x] - 24 x^2 y''[x] + 48 x y'[x] - 48 y[x]]

d[  Function[x,x^n ) ]  =   6 (-4 + n) (-2 + n) (-1 + n) x^n



It follows that $$n=1,2,4$$ solve and by linearity any linear combination, too

   d[x |-> (a x + b x^2 + c x^4)] = 0