.Hey, first time post, I found this place by Googling "Math Help Forum". Sorry for not knowing the script to write out math eqn's, so here goes.
I'm trying to figure out the following ODE:
y' - 6tan(3x)*y + 2cos(5x) = 0
I'm 99% sure it's a Linear ODE so I rewrote it as:
y' - 6tan(3x)*y = -2cos(5x)
Found an integrating factor mu = e^(2ln(cos(3x)) = (cos(3x))^2 Mr F says: Correct.
However this does not leave me the result of a product rule. Mr F says: Yes it does (see main reply).
y'*(cos(3x))^2 - 6sin(3x)*cos(3x)*y = (-2cos(5x))^3
I tried dividing by cos(3x) which gave me
cos(3x)*y' - 6sin(3x)*y = (-2cos(5x))^2
Which looks like it should be the result of a product derivative, however at this point I know something is wrong, as either I need to have 2cos(3x) or -3sin(3x) for things to work out. Where am I going wrong?
So there's no trouble here. Go ahead and continue the process.