how do i find the soln to xv''-2v'+2xv'=0 from linear homogenous second order de
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
i havent learned lambda yet. so do you understand how to do it without it?
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
you are too smart for the answer i am looking for. thanx though
I dont know any different way of solving this :(
but this is one seems really hard as there is v', v'' and x in it
the original problem was..... xy''-2(x+1)(y'+4y=0 with y1(x)=e^2x find the general soln
aha I see, reduction of order
still i dont see how that helps, wait a sec let me think .-)
what is that y1(x), what does the 1 stand for?
the first solution
try to find the second soln y2x to create the general soln
ill ask my prof. thanks
yeah I am not sure
nice one nikvist! what did you use to make that, just pdf?
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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