- #1

chwala

Gold Member

- 2,662

- 352

- Homework Statement
- My own question (set by me):

##4x^2 \dfrac{d^2u}{dx^2} +12x\dfrac{du}{dx} -4u=0##

- Relevant Equations
- Cauchy-Euler equations

Going through ode's,

My lines,

Let solution be of the form, ##u =x^s##

...

Characteristic equation is;

##4(s(s-1)+12s-4=0##

##4s^2+8s-4=0##

##s_1=\sqrt 2 -1## and ##s_2 = -(\sqrt 2+1)##

thus the two linearly indepedent solutions are,

##u_1(x)= x^{\sqrt 2 -1}## and ##u_2 = \dfrac{1}{x^{\sqrt 2+1}}##

i think that's fine but of course any insight is welcome.... i find such problems to be convenient ie as long as you know the steps you're good to go.

Supposing i had,

##4x^4 \dfrac{d^2u}{dx^2} +12x^2\dfrac{du}{dx} -4u=0## would one make use of another variable? say let ##v=x^2?##

What about,

##4x^5 \dfrac{d^2u}{dx^2} +12x^2\dfrac{du}{dx} -4u=0##

Cheers

My lines,

Let solution be of the form, ##u =x^s##

...

Characteristic equation is;

##4(s(s-1)+12s-4=0##

##4s^2+8s-4=0##

##s_1=\sqrt 2 -1## and ##s_2 = -(\sqrt 2+1)##

thus the two linearly indepedent solutions are,

##u_1(x)= x^{\sqrt 2 -1}## and ##u_2 = \dfrac{1}{x^{\sqrt 2+1}}##

i think that's fine but of course any insight is welcome.... i find such problems to be convenient ie as long as you know the steps you're good to go.

Supposing i had,

##4x^4 \dfrac{d^2u}{dx^2} +12x^2\dfrac{du}{dx} -4u=0## would one make use of another variable? say let ##v=x^2?##

What about,

##4x^5 \dfrac{d^2u}{dx^2} +12x^2\dfrac{du}{dx} -4u=0##

Cheers

Last edited: