Question

Cauchy-Euler Equation
I really dont know how to do this anyone care to explain in detail?
xy'''+2y"=0

Answer 1

if you let \[u=y \prime \prime\] the DE is transformed to \[xu \prime+2u=0\]
a DE that can be solved by separation

Answer 2

i dont quite understand my teacher used like P(D)y = xD^3+2D2 = 0 then multiplied \[x^2\] so it becomes \[x^2D^3+2D^2 = 0 \] then somehow separated into \[xD(D-1)(D-2) + 2D(D-1) = 0\] i dont understand this part

Answer 3

\[x^3D^3+2x^2D^2\]

Answer 4

could you solve it if it was:

xy' + 2y = 0 ??

Answer 5

im not familiar with the D as an operator so i cant attest to your teacher method

Answer 6

hm... would it be x(D+2)y = 0?

Answer 7

let x = e^t

Answer 8

im not sure... this is confusing...

Answer 9

http://www.wolframalpha.com/input/?i=xy%27%27%27%2B2y%27%27%3D0

i keep mismathing it along the way, but this at least gives us something to dbl chk against