Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If someone could help me with the proposed change of variables, it would be greatly appreciated. Consider Euler's equation:

$$z^2w'' + \alpha zw' + \beta w = 0$$

where $w$ is a function of $z$ and $\alpha, \beta$ are constants. How would the change of variables $t =\ln(z)$ transform this equation?

share|cite|improve this question

1 Answer 1

up vote 1 down vote accepted

This is calculus, so we calculate. Note that $$\frac{dw}{dz}=\frac{dw}{dt}\frac{dt}{dz}=\frac{dw}{dt}\frac{1}{z}.\tag{$1$}$$

For the second derivative, we need to differentiate $\frac{dw}{dt}\frac{1}{z}$ with respect to $z$. Use the Product Rule. We get $$\frac{dw}{dt}\left(-\frac{1}{z^2}\right) +\frac{1}{z}\frac{d}{dz}\left(\frac{dw}{dt}\right).$$

For the unfinished second part, we get $$\frac{1}{z}\frac{d^2w}{dt^2}\frac{1}{z},$$ by the same sort of Chain Rule calculation that we used for $(1)$.

Put things together. We get $$\frac{d^2 w}{dz^2}=\frac{1}{z^2}\left(-\frac{dw}{dt}+\frac{d^2w}{dt^2}\right).\tag{$2$}$$

Now substitute for $\frac{d^2w}{dz^2}$ and $\frac{dw}{dz}$ in the original DE. The $z^2$'s cancel in the first term, and the $z$'s cancel in the second term. We end up with a very tractable DE. This was the whole purpose for the change of variable.

share|cite|improve this answer
Thank you very much! – user44069 Feb 19 '13 at 6:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.