Question

Find the general solution of the nonhomogeneous equation

y''+4y=sec 2x

Answer 1

find the solution of homogeneous equation first.

Answer 2

by just ignoring the sec2x or you saying move it to the other side?

Answer 3

yes by ignoring it ...
y''+4y=0 , find it's solution

Answer 4

okay so r=+/2i so alpha is zero and beta is 2

so y = C1cos2x + C2sin2x

Answer 5

okay ... your y1 = sin2x and y2=cos2x
use variation of parameters as before.

Answer 6

to find the particular solution ... or try using
http://en.wikipedia.org/wiki/Reduction_of_order
i wonder if you can solve using http://en.wikipedia.org/wiki/Annihilator_method

Answer 7

does it matter is y1 is sin 2x or cos 2x, does it make a difference which is labeled what?

Answer 8

no it doesn't ... just be consistent with what you choose initially.

Answer 9

does this simplify, or should I say what does it simplify too?

2(cos2x)(cos2x) + 2(sin2x)(sin2x)
2( (cos2x)^2 + (sin2x)^2)

is that sec?? I have trouble remembering the identities?

Answer 10

seems that the wonskian is 2 ... you should feel lucky.

Answer 11

ohhh, thats right.. cos 2x squared plus sin 2x squared is = 1. that does make it easier.

Answer 12

u(x) is horrible... I am not even sure how to begin to simplify z(x)

\[z(x)=\frac{ \ln(|(\sin(2x))^{2}-1|) }{ 8 }(\cos2x)+\frac{ x }{ 2 }(\sin2x)\]

Answer 13

hmm ... did you really get that?? 1 - sin^2(2x) or ... the opposite?

Answer 14

well i typed it into the integral calculator, and that is what it gave me.

I typed in integral of -sin(2x)sec(2x)/2

Answer 15

you did the opposite ... it was 1 - sin^2(2x) = cos^2(2x) ... get the square outside of log.
http://www.wolframalpha.com/input/?i=integrate+-sin%282x%29sec%282x%29%2F2

Answer 16

I typed in exactly what you did into the WA, but for some reason it gave me that crazy answer... but I see it know... thank you again for your help.