Question

Consider the equation y'' -xy' + x^2*y = 0. Find a solution y1 in the form of y = x^r, where r is a constant.

Answer 1

So substitute y = x^r into the equation and you'll get an equation in r you can solve.

Answer 2

y=Ax^r
find y' and y''
and plug into the equation

Answer 3

Try it. It's surprisingly easy.

Answer 4

awesome thanks

Answer 5

i don't see how it can be a solution unless i missing something

Answer 6

Yeah I'm not really getting anything out of this substitution either. I don't think there are any solutions that are real.

Answer 7

well i get
\[x^r \cdot [r(r-1)x^{-2}-r+x^2]=0 \]
x^r=0 when x=0
-----------
\[r(r-1)x^{-2}-r+x^2=0\] will not give us a constant r

Answer 8

we don't want x=0 since we are looking for a solution in the form y=x^r
\[r(r-1)-rx^2+x^4=0\]
\[r^2-r-rx^2+x^4=0 => r^2+r(-1-x^2)+x^4=0\]

Answer 9

Oh wait. Have you written down the equation correctly?

Answer 10

there is probably a typo in the question

Answer 11

yes -- I think so too.

Answer 12

move the x^2 to y''' prob

Answer 13

y''?

Answer 14

y''

Answer 15

lol

Answer 16

too quick with the primes ;)

Answer 17

yes x^2y'' - 2xy' + y = 0 would make sense.

Answer 18

That's how its written, I thought it could have been a typo too but put it in case anyone saw something that I didn't. The prof is notorious for having a few typos in each assignment and not telling us till earlier on the day its due, so I'm guessing that's the case here as well.

Answer 19

http://www.wolframalpha.com/input/?i=y%27%27+-x%2Ay%27+%2B+x^2%2Ay+%3D+0&t=ietb01

Answer 20

what you have is a nasty diff EQ

Answer 21

Yes, otherwise the solution will very messy, as Zarkon just demonstrated.

Answer 22

I'd assume it is the form I wrote above. Then it really IS easy!

Answer 23

i would say there is no such solution in the form y=x^r for the dq given
and then you could show there is a solution of the form to what james wrote (just in case that is what the prof meant)