# Thread: O.D.E Help

1. ## O.D.E Help

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

However this does not leave me the result of a product rule.

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?

2. Originally Posted by BigC
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?
$\frac{d}{dx} \left[ y \cos^2 (3x)\right] = \cos^2 (3x) \frac{dy}{dx} - 6 \sin (3x) \cos (3x) y$.

So there's no trouble here. Go ahead and continue the process.

3. I guess I just need to practice and re-memorize my rules of differentiation. I didn't even remember the derivative of cos(x)^2
Thanks for the help