Question

For what values of p is y = x^p a solution to the differential equation x^2y'' - 3xy' - 5y = 0 ?

Answer 1

Anyone?

Answer 2

y=x^p
y'=px^{p-1}
y''=p(p-1)x^{p-2}

x^2(p(p-1)x^{p-2})-3x(px^{p-1})-5(x^p)
=p(p-1)x^p-3x^p -5x^p
=(p^2-p)x^p-8x^p
=(x^p)(p^2-p-8)

Answer 3

do p^2-p-8=0 to find your p's

Answer 4

let y=x^r
y'=rx^r-1
y''=r(r-1)x^r-2
x^r(r-1)-3x^r(r)-5x^r=0
x^r(r^2-r-3r-5)=0
x^r(r^2-4r-5)=0
x^r((r-5)(r+1))=0
(r-5)(r+1)=0

therefore the real roots are y(x)= c1x^5+c2x^-1

Answer 5

oops i missed something

Answer 6

did you check mine myin

Answer 7

nut got it

Answer 8

i forgot my p in the 3x^p term

Answer 9

suppose to be 3px^p

Answer 10

gj nut

Answer 11

ok ok thanks myin ;)

Answer 12

Thanks a lot bnut.

Answer 13

And thank you too myininaya.

Answer 14

np