Question

For what values of k0 and k1 does the initial value problem x^2*y"-4xy'+6y=0, y(0)=k0, y'(0)=k1 have a solution?

Answer 1

@pooja195 @Michele_Laino @campbell_st @uri @timo86m

Answer 2

@mathstudent55 @whpalmer4 @just_one_last_goodbye @sammixboo @shifuyanli

Answer 3

@amistre64 @freckles @dan815 @KendrickLamar2014 @whpalmer4 @sammixboo

Answer 4

does laplace work for this?

Answer 5

dividing by x^2 would mean that x=0 is a sore subject ...

Answer 6

x^2*y"-4xy'+6y=0

let y=x^n

x^2*n(n-1)x^(n-2)-4x nx^(n-1)+6x^n=0

n(n-1)x^n-4 nx^n+6x^n=0

x^n [ n(n-1) -4n +6]=0

n^2-5n+6=0

(n-3)(n-2)=0, n=3 or 2

y=ax^3 + bx^2 is our general solution

Answer 7

Now what? And then?

Answer 8

determine of y(0) and y'(0) have a solution set ... of course.

Answer 9

I got it! Thanks for the help!