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DanJS
 one year ago
Solve:
y'' + y = csc(x) ; 0<x<pi
DanJS
 one year ago
Solve: y'' + y = csc(x) ; 0<x<pi

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.3characteristic equation is??

Loser66
 one year ago
Best ResponseYou've already chosen the best response.3nope, solve it by characteristic equation.

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1For the homogeneous solution to L(y) = 0 characteristic eq is.. r^2 + 1 = 0 r = + or  i \[y _{h} = c _{1}e^{i * x} + c _{2}e^{i * x}\]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1been a few years, bear with me, looking at the book. lol

Loser66
 one year ago
Best ResponseYou've already chosen the best response.3is it not that if \(r=\pm i\) , then \(y_h= C_1 cos (x) + C_2 sin(x)\)??

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1so you can say...from Euler

Loser66
 one year ago
Best ResponseYou've already chosen the best response.3ok, now solve for partial part.

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1power series expansion of e^(ix)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1Particular solution to L(y) = f(x), ... hmm

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1looking at variation of parameter technique

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1\[y _{p} = g _{1}u _{1} + g _{2}u _{2}\] so, u1 = cos(x) u2 = sin(x)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1and \[g _{1}^{~'}*u _{1} + g _{2}^{~'}*u _{2} = 0\] \[g_{1}^{'}*u _{1}^{'} + g_{2}^{'}*u _{2}^{'} = f(x)\] so just put in u1 and u2, and solve g1 and g2 prime ?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1then integrate to get the g1 and g2 for the particular solution...?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1i remember memorizing those 2 lines , forgot where they come from, the = 0 and = f(x) , things

Loser66
 one year ago
Best ResponseYou've already chosen the best response.3I am sorry, I am not familiar with this method. I use Wronskian to solve it. :) @ganeshie8 Please.

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1ah right, i did not do that one

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1so i got \[g _{2} = \ln(\sin(x)) ~~~so~~~~g _{1} = x\] then, \[y _{p} = x*\cos(x) + \ln(\sin x)*\sin(x)\]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1u get that with your method?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1it has to be right, it cancelled nicely, and is a book 'nice' problem.

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1\[y _{g} = y _{h} + y _{p}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0looks good, wolfram agrees :) http://www.wolframalpha.com/input/?i=solve+y%27%27+%2B+y+%3D+csc%28x%29++++++++++

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1cool, trying to do a couple of each solution technique to review, i forget a F ton
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