How to eliminate the first order term from a second order ODE

Authors will often say that “without loss of generality” they will assume that a differential equation has no first order derivative term. They’ll explain that there’s no need to consider

y'' + p(x) y' + q(x) y = 0

because a change of variables can turn the above equation into one of the form

y'' + q(x) y = 0

While this is true, the change of variables is seldom spelled out. I’ll give the change of variables explicitly here in case this is helpful to someone in the future. Define u(x) and r(x) by

u(x) = \exp\left( \frac{1}{2} \int^x p(t)\,dt\right ) y(x)


r(x) = q(x) - \frac{1}{2}p'(x) - \frac{1}{4}p(x)^2

With this change of variables

u'' + r(x) u = 0

Proof: Calculate u” + ru and use the fact that y satisfies the original differential equation. The calculation is tedious but routine.

Example: Suppose we start with

xy'' + y' + x^3 y = 0

Then dividing by x we get

y'' + \frac{1}{x}y' + x^2 y = 0

Now applying the change of variables above gives

u'' + \left(x^2 + \frac{1}{4x^2}\right)u = 0
and our original y is u / √ x.

Update: Here is a new post generalizing the content of this post to higher-order ODEs.

3 thoughts on “How to eliminate the first order term from a second order ODE

  1. Rafael Josẽ Iorio, Jr.

    Very useful. I am a mathematician too.. And we have this perversion of making things obscure to students and for those who read our articles. Not I. But most do. I think this information is very important for students to learn about second order equations. Congratulations and keep on with the good works. Rafael Josẽ Iorio, Jr.

  2. What about for a general n-th order linear ODE? It’s always possible to redefine our function such that the (n-1) derivative term dies out (they say), but I’m having trouble generalizing to this case…

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