Question

For which value(s) of m is y=x^m a solution to the DE x^2y''-y=0?

Answer 1

u can first differentiate y, twice and find y''
what u get?

Answer 2

then check whether x^2 y'' - y comes out to be 0, if its 0, then its the solution

Answer 3

I tried dat..let me try again
...by the way am asked for the specic m values...

Answer 4

oh, yes, u will get a quadratic in m when u equate that to 0, and u will get specific values of m

Answer 5

did u get y" as m(m-1)x^(m-2) = m(m-1)x^m/x^2......?

Answer 6

I got x^2((m-1)(m)x^(m-2))-x^m=0........After differentiation and substitution!!

Answer 7

then add the exponents of x^2 and x^(m-2)

Answer 8

=x^m

Answer 9

x^(m-2) = x^m/x^2
so
x^2 * x^(m-2) = x^m

Answer 10

then take x^m common

Answer 11

so finally did u get
m^2-m-1=0
can u solve this ?

Answer 12

=x^m
x^(m-2) = x^m/x^2
so
x^2 * x^(m-2) = x^m

Answer 13

i think that is true answer

Answer 14

but u cannot solve for m to make the equation =0 canu?

Answer 15

u can solve his
m^2-m-1=0
or u have some other doubt ?

Answer 16

cool...so it's not a whole number
I see...

Answer 17

yup, its not a whole number.

Answer 18

u want a medal?lol

Answer 19

if u get the right answer.....

Answer 20

:)

Answer 21

thank u :)