Question

Need help solving this differential equation...

Answer 1

\[2y + \frac{dy}{dx}\tan(2x) =\cos(2x)\]

Answer 2

\[\frac{dy}{dx} = \frac{\cos(2x) -2y}{\tan(2x)}\]

Answer 3

how do i integrate the right side?

Answer 4

okay some guy delete his message so that probably is confusing. I think i need to use the integration factor method

Answer 5

because the original equation is
\[\frac{dy}{dx} + \frac{2}{\tan(2x)} \times y = \cos(2x)\]

Answer 6

so the integrating factor will be, \[\exp(\int\limits_{}^{}\frac{2}{\tan2x}dx)\]

Answer 7

is that right?

Answer 8

http://mathworld.wolfram.com/First-OrderOrdinaryDifferentialEquation.html

Answer 9

so that is the integrating factor im looking for?

Answer 10

\[\exp(\int\limits\limits_{}^{}\frac{2}{\tan2x}dx) = 2 \sin(2x)\]
is that right?

Answer 11

yes

Answer 12

so that turns the problem into
\[\sin(2x)y = \int\limits_{}^{} \cos(2x)\sin(2x) dx\]

Answer 13

no... \[\sin(2x)y = \int\limits\limits_{}^{} \frac{\cos(2x)\sin(2x)}{\tan(2x)} dx\]

Answer 14

because i have to multiply everything by tan at the start.

Answer 15

and this is...
\[\sin(2x)y = \int\limits_{}^{}\cos^2(2x) dx\]

is that right?

Answer 16

\[y'+2cotan(2x)y=\cos(2x)cotan(2x)\]

\[\large y'e^{\int(2cotan(2x))dx}+2cotan(2x)e^{\int(2cotan(2x))dx}y=\cos(2x)cotan(2x)e^{\int(2cotan(2x))dx}\]
\[\large (y \cdot e^{\int(2cotan(2x))dx})'=\cos(2x)cotan(2x)e^{\int(2cotan(2x))dx}\]
\[\large y \cdot e^{\int(2cotan(2x))dx}=\int \cos(2x)cotan(2x)e^{\int(2cotan(2x))dx} dx \]

Answer 17

If it helps

Answer 18

\[y e^{f(x)}=\int\limits \frac{\cos^2(2x)}{\sin(2x)}e^{f(x)}dx\]
Maybe use Euler's formula to make the integration nicer

Answer 19

Maybe just kill myself :(

Answer 20

not certain why this is so bloody hard

Answer 21

Did my equations help or did you know that already?

Answer 22

so your turning 1/tan into cotan, im not certain how that helps henpen :(