DanJS
  • DanJS
Solve: y'' + y = csc(x) ; 0
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
DanJS
  • DanJS
general soln
Loser66
  • Loser66
characteristic equation is??
DanJS
  • DanJS
variation of parameters

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Loser66
  • Loser66
nope, solve it by characteristic equation.
DanJS
  • DanJS
For the homogeneous solution to L(y) = 0 characteristic eq is.. r^2 + 1 = 0 r = + or - i \[y _{h} = c _{1}e^{i * x} + c _{2}e^{-i * x}\]
DanJS
  • DanJS
been a few years, bear with me, looking at the book. lol
Loser66
  • Loser66
is it not that if \(r=\pm i\) , then \(y_h= C_1 cos (x) + C_2 sin(x)\)??
DanJS
  • DanJS
so you can say...from Euler
DanJS
  • DanJS
right
Loser66
  • Loser66
ok, now solve for partial part.
DanJS
  • DanJS
power series expansion of e^(ix)
DanJS
  • DanJS
Particular solution to L(y) = f(x), ... hmm
DanJS
  • DanJS
minute
DanJS
  • DanJS
looking at variation of parameter technique
DanJS
  • DanJS
\[y _{p} = g _{1}u _{1} + g _{2}u _{2}\] so, u1 = cos(x) u2 = sin(x)
DanJS
  • 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 ?
DanJS
  • DanJS
then integrate to get the g1 and g2 for the particular solution...?
DanJS
  • DanJS
i remember memorizing those 2 lines , forgot where they come from, the = 0 and = f(x) , things
Loser66
  • Loser66
I am sorry, I am not familiar with this method. I use Wronskian to solve it. :) @ganeshie8 Please.
DanJS
  • DanJS
ah right, i did not do that one
DanJS
  • DanJS
so i got \[g _{2} = \ln(\sin(x)) ~~~so~~~~g _{1} = -x\] then, \[y _{p} = -x*\cos(x) + \ln(\sin x)*\sin(x)\]
DanJS
  • DanJS
u get that with your method?
DanJS
  • DanJS
it has to be right, it cancelled nicely, and is a book 'nice' problem.
DanJS
  • DanJS
\[y _{g} = y _{h} + y _{p}\]
ganeshie8
  • ganeshie8
looks good, wolfram agrees :) http://www.wolframalpha.com/input/?i=solve+y%27%27+%2B+y+%3D+csc%28x%29++++++++++
DanJS
  • DanJS
cool, trying to do a couple of each solution technique to review, i forget a F ton

Looking for something else?

Not the answer you are looking for? Search for more explanations.