# Euler-cauchy equation

how to find the general solution for the equation
(10x^2D^2-20xD+22.4I)y=0...no idea to find it by using euler-method...help..sos

HallsofIvy
Homework Helper
Unfortunately, Euler's name is attached to to many things!

That is an "Euler" or "Euler-type" equation. One way of solving it is to try a solution of the form $y= x^r$ for some number r. Then $y'= rx^{r-1}$ and $y''= r(r-1)x^{r-2}$. Putting those into the equation gives
$(10r(r-1)+ 20r+ 22.4)x^r= 0$ and, since $x^r$ is not always 0,
$10r(r- 1)+ 20r+ 22.4= 10r^2+ 10r+ 22.4= 0$

That will have complex solutions so you would have to interpret $x^{a+ bi}$ in terms of sine and cosine.

Another way of doing that would be to make the change of variable x= ln(t) which converts the "Euler-type" equation to an equation with constant coefficents having the same characteristic equation.

Or, since you say "Euler method" are you referring Euler's numerical method for solving equations? Strictly speaking that applies to first order equations but you can convert this second order equation to a pair of first order equations:

Let y= Dy so that $D^2 y= Du$ any your equation becomes
$x^2Du- 20xDy+ 22.4y= 0$
and $Dy= u$
You can set up an "double" Euler-method solver that, given the values of u and y at a each step, solves Dy= u for y and then puts both of those values into the other equation to solve for y.

but the answer given to the equation is C1x^1.4 + C2X^1.6,its no complex solutions....and i just learn the euler cauchy equation only..