Question

Need Help to solve this differential equation
\[\frac{dy}{dx} + \frac{2}{\tan(2x)} \times y = \cos(2x)\]

Answer 1

its integrating factor method. so the integrating factor here will be
\[\sin(2x)\]

Answer 2

is that right?

Answer 3

yes

Answer 4

so \[\sin(2x)y = \int\limits{}^{}\frac{\cos(2x)\sin(2x)}{\tan(2x)}\]

Answer 5

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

Answer 6

where did the tan(2x) come from?

Answer 7

sorry i typed it wrong.

Answer 8

the very first equation was
\[2y + \frac{dy}{dx}\tan(2x) = \cos(2x)\]

so i divided by tan to get it linear

Answer 9

ok, then so far so good
you should simplify \[\frac{\sin(2x)\cos(2x)}{\tan(2x)}\]before integrating though
btw my connection is horrible, sorry if I'm slow to reply.

Answer 10

\[\int\limits_{}^{} \cos^2(2x) dx\]

Answer 11

yep

Answer 12

not sure how to do this.

Answer 13

cos(4x) = 2cos^2(2x) - 1

Answer 14

so \[1/2 \int\limits_{}^{} 2\cos^2(2x) - 1 + 1 = 1/2 \int\limits_{}^{} \cos(4x) + 1\]

Answer 15

is that right?

Answer 16

yes

Answer 17

btw I know @Coolsector 's formula better as\[\cos^2x=\frac12(1+\cos(2x))\]just so you know where it came from...

Answer 18

\[y = \frac{\frac{1}{8} \sin(4x) + x}{\sin(2x)}\] ?

Answer 19

so using yours Turing would give:
\[\int\limits_{}^{} \frac{1}{2} + \frac{1}{2}\cos(2x) = \frac{1}{2}x + \frac{1}{4}\sin(2x)\]

Answer 20

sorry i was doing cos^2(x) not cos^2(2x)

Answer 21

\[\int\limits\limits_{}^{} \frac{1}{2} + \frac{1}{2}\cos(2x) = \frac{1}{2}x + \frac{1}{8}\sin(2x)\]

Answer 22

is that right?

Answer 23

no, the argument after using the trig formula is 4x
\[\cos^2(x)=\frac12(1+\cos(2x))\implies\cos^2(2x)=\frac12(1-\cos(4x))\]so you and coolsector had it right
I was just trying to make a point, sorry if I confused you

Answer 24

yeah sorry i didnt get that.

Answer 25

typo, that should be a + in the second equation above...

Answer 26

\[y = \frac{ \sin(4x) }{8\sin(2x)} + \frac{x}{8}\] ?

Answer 27

ahh i see Turing. So is that the solution for y? (plus a constant)

Answer 28

\[\int\frac12+\frac12(\cos(4x))dx=\frac12x+\frac18\sin(4x)+C=y\sin(2x)\]differential equations is when you find out how incredibly important the integration constant C is; it will *completely* ruin your answer if you neglect it.
solve for y

Answer 29

always forget C :(

Answer 30

the problem is an initial value problem y(pi/4) = 1

Answer 31

ok, first solve for y to get the general
you don't have to do it that way first, but it's a good habit

Answer 32

so sin(4pi/4) = 0 and sin(2pi/4) = 1 which gives
\[\frac{\pi}{8} = y\]

Answer 33

no sorry , pi/8 = -C

Answer 34

that's right :)
your explicit solution really should utilize more trig identities as well

Answer 35

it does seem quite complicated. Which identities do you think?

Answer 36

\[\frac12x+\frac18\sin(4x)+C=y\sin(2x)\]using the double angle formula for sine\[\sin(2x)=2\sin x\cos x\]we get\[\frac12x+\frac14\sin(2x)\cos(2x)+C=y\sin (2x)\]then just divide by sin(2x) and remember that 1/sinx=cscx to make it look nicer

Answer 37

Awesome! Im useless at remembering identities too

Answer 38

thanks.

Answer 39

no problem :)