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OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

I dont know any different way of solving this :(

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

substitute v=y'

OpenStudy (anonymous):

aha I see, reduction of order

OpenStudy (anonymous):

yeah :)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

the first solution

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ill ask my prof. thanks

OpenStudy (anonymous):

yeah I am not sure

OpenStudy (nikvist):

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.

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