Question

Solve:

y'' + y = csc(x) ; 0 10 years ago

Answer 1

general soln

Answer 2

characteristic equation is??

Answer 3

variation of parameters

Answer 4

nope, solve it by characteristic equation.

Answer 5

For the homogeneous solution to L(y) = 0
characteristic eq is..
r^2 + 1 = 0

r = + or - i

\[y _{h} = c _{1}e^{i * x} + c _{2}e^{-i * x}\]

Answer 6

been a few years, bear with me, looking at the book. lol

Answer 7

is it not that if \(r=\pm i\) , then \(y_h= C_1 cos (x) + C_2 sin(x)\)??

Answer 8

so you can say...from Euler

Answer 9

right

Answer 10

ok, now solve for partial part.

Answer 11

power series expansion of e^(ix)

Answer 12

Particular solution to L(y) = f(x), ... hmm

Answer 13

minute

Answer 14

looking at variation of parameter technique

Answer 15

\[y _{p} = g _{1}u _{1} + g _{2}u _{2}\]

so,
u1 = cos(x)
u2 = sin(x)

Answer 16

and

\[g _{1}^{~'}*u _{1} + g _{2}^{~'}*u _{2} = 0\]
\[g_{1}^{'}*u _{1}^{'} + g_{2}^{'}*u _{2}^{'} = f(x)\]


so just put in u1 and u2, and solve g1 and g2 prime ?

Answer 17

then integrate to get the g1 and g2 for the particular solution...?

Answer 18

i remember memorizing those 2 lines , forgot where they come from, the = 0 and = f(x) , things

Answer 19

I am sorry, I am not familiar with this method. I use Wronskian to solve it. :)
@ganeshie8 Please.

Answer 20

ah right, i did not do that one

Answer 21

so i got \[g _{2} = \ln(\sin(x)) ~~~so~~~~g _{1} = -x\]

then,

\[y _{p} = -x*\cos(x) + \ln(\sin x)*\sin(x)\]

Answer 22

u get that with your method?

Answer 23

it has to be right, it cancelled nicely, and is a book 'nice' problem.

Answer 24

\[y _{g} = y _{h} + y _{p}\]

Answer 25

looks good, wolfram agrees :)
http://www.wolframalpha.com/input/?i=solve+y%27%27+%2B+y+%3D+csc%28x%29++++++++++

Answer 26

cool, trying to do a couple of each solution technique to review, i forget a F ton