Equi Potential Equations or Euler Type Equations

The main thing you need to know about "Euler-type" equations is that there is a "one-to-one" correspondence between them and equations with constant coefficients.

Just as you can "look for" solutions to equations with constant coefficients using $e^{rx}$ (even though solutions may not be exponential), so you can "look for" solutions to "Euler-type" equations using $x^r$ .

That's because the change of variable x= ln(t) will change an "Euler-type" equation in variable t to an equation with constant coefficients in variable t.

For example, $at^2 \frac{d^2y}{dt^2}+ bt\frac{dy}{dt}+ cy= 0$ is an "Euler-type" equation in variable t. If we let x= ln(t), then $\frac{dy}{dt}= \frac{dy}{dx}\frac{dx}{dt}= \frac{1}{t} \frac{dy}{dx}$ .

Further $\frac{d^2y}{dt^2}= \frac{d}{dt}\left(\frac{1}{t} \frac{dy}{dx}\right)$ $= \frac{1}{t}\frac{d}{dt}\left(\frac{1}{t}\frac{dy}{ dx}\right)= \frac{1}{x}\left(\frac{1}{t}\frac{d^2y}{dx}^2- \frac{1}{t^2}\frac{dy}{dx}\right)$ .

So $at^2\frac{d^2y}{dt^2}+ bt\frac{dy}{dt}+ cy$ $= a\left(\frac{d^2y}{dx^2}- \frac{dy}{dx}\right)+ b\frac{dy}{dx}+ cy$ $= a\frac{d^2y}{dx^2}+ (b- a)\frac{dy}{dx}+ cy= 0$ .