- #1
chwala
Gold Member
- 2,703
- 369
- 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: