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

1. DanJS

general soln

2. Loser66

characteristic equation is??

3. DanJS

variation of parameters

4. Loser66

nope, solve it by characteristic equation.

5. DanJS

For the homogeneous solution to L(y) = 0 characteristic eq is.. r^2 + 1 = 0 r = + or - i $y _{h} = c _{1}e^{i * x} + c _{2}e^{-i * x}$

6. DanJS

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

7. Loser66

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

8. DanJS

so you can say...from Euler

9. DanJS

right

10. Loser66

ok, now solve for partial part.

11. DanJS

power series expansion of e^(ix)

12. DanJS

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

13. DanJS

minute

14. DanJS

looking at variation of parameter technique

15. DanJS

$y _{p} = g _{1}u _{1} + g _{2}u _{2}$ so, u1 = cos(x) u2 = sin(x)

16. DanJS

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 ?

17. DanJS

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

18. DanJS

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

19. Loser66

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

20. DanJS

ah right, i did not do that one

21. DanJS

so i got $g _{2} = \ln(\sin(x)) ~~~so~~~~g _{1} = -x$ then, $y _{p} = -x*\cos(x) + \ln(\sin x)*\sin(x)$

22. DanJS

u get that with your method?

23. DanJS

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

24. DanJS

$y _{g} = y _{h} + y _{p}$

25. ganeshie8

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

26. DanJS

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