Question

Find a general solution for
2y''+18y=6tan(3t)

Should be y(t)=Something
Come on... anyone doing diferential calculus out there?

Answer 1

http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

Answer 2

So far i have y(t)=C1*cos(3t)+C2*sen(3t)+sen(3t)cos(3t)/3 + Y2p

missing that second particular solution, but do i have everything ok so far?

Answer 3

@dumbcow hahah I am reading off that site right now paul's online notes for the win.

Answer 4

look at example 1
http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx

Answer 5

Lol, @dumbcow thanks ;)

Answer 6

just trying to get that \[\int\limits_{}Sec(3t)dt\]

Answer 7

the guy solves that out of nowhere...

Answer 8

im no expert at these btw

that integral can be solved by multiplying top/bottom by sec + tan
then use substitution:
u = sec + tan
du = sectan + sec^2

Answer 9

Yeah!! I guessed the correct 1st step haha!! Sorry small accomplishment I know. Just started learning 1st order last night so it feels good to get at least something out of the problem.

Answer 10

I think that was meant for someone else @dumbcow

Answer 11

Small my retrice that was really fine, mult by sec+tan...

Answer 12

oh, lol, meant that sec + tan to dumbcow

Answer 13

\[\int\limits_{}^{} \sec(3t)*\frac{\sec(3t)+\tan(3t)}{\sec(3t)+\tan(3t)} dt = \int\limits_{}^{}\frac{\sec^{2}(3t)+\sec(3t) \tan(3t)}{\sec(3t)+\tan(3t)}dt=\frac{1}{3}\int\limits_{}^{}\frac{du}{u}\]
given that u = sec(3t) +tan(3t)

Answer 14

hum.. i'm solving that right now, shouldn't du be different, i mean it is 3t so that 3 will be poping up couple times

Answer 15

true, but then you factor the 3 out .....thats where the 1/3 comes from

du = 3sectan + 3sec^2 = 3(sec^2 + sectan) dt

Answer 16

d(Sec(3t))/dt = 3Sec2(3t)sen(3t), right?

Answer 17

nvmd

Answer 18

same thing in diff form :)

Answer 19

ok, got it ;)