Question

how do i find the soln to xv''-2v'+2xv'=0 from linear homogenous second order de

Answer 1

you use the auxiliary equation, so you replace v'' by lambda^2 v' by lambda and v by 1
than solve the quadratic equation you will get x1, x2 values.
there are 3 different possibilities
1, if x1 and x2 are 2 different real numbers (like 1 and 3) than the solution will be
Ae^(x1*x)+Be^(x2*x)
where A and B are constants
2, if x1=x2 the solution is
(A+Bx)e^(x1*x)
3, if the roots are imaginary the solution will be
Acosx1x + Bsinx2 x

Answer 2

i havent learned lambda yet. so do you understand how to do it without it?

Answer 3

sorry the last bit is wrong, if it is complex than the solution is
e^rx(Acosx1x+Bcosx2x) where r is the real part of the complex number

Answer 4

you are too smart for the answer i am looking for. thanx though

Answer 5

I dont know any different way of solving this :(

Answer 6

but this is one seems really hard as there is v', v'' and x in it

Answer 7

the original problem was..... xy''-2(x+1)(y'+4y=0 with y1(x)=e^2x find the general soln

Answer 8

substitute v=y'

Answer 9

aha I see, reduction of order

Answer 10

yeah :)

Answer 11

still i dont see how that helps, wait a sec let me think .-)

Answer 12

what is that y1(x), what does the 1 stand for?

Answer 13

the first solution

Answer 14

try to find the second soln y2x to create the general soln

Answer 15

ill ask my prof. thanks

Answer 16

yeah I am not sure

Answer 17

nice one nikvist! what did you use to make that, just pdf?

Answer 18

thank you very much! i asked the prof and she gave me the same answer :) oh and there is an attach file buttion at the bottom of the comment to add something.