Anyone particularly hot on differential equations?

Arcade Fire

Arcade Fire

Arcade Fire

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Help me with my research! The differential equation

dy/dx = N cos(y)

where N is a constant, has come up, and I can't remember how to solve it. Anyone who helps me can get, I dunno, prestige or something.
 
Make it

dy/ cos (y) = dx N

Then integrate both sides?

1 / cos (y) will go to 1 / 2 (sin (y))^2 or something, I can't remember integration of cos and sin, google for it. Other side is easy.

I don't know if that's what you're after, or if it's even correct, but there's my input!
 
Carzy said:
Make it

dy/ cos (y) = dx N

Then integrate both sides?

1 / cos (y) will go to 1 / 2 (sin (y))^2 or something, I can't remember integration of cos and sin, google for it. Other side is easy.
Click to expand...
Nah, differentiating 1/(sin y)^2 gives -2 cos y/(sin y)^3, not 1/cos y

:(
 
Arcade Fire said:
Nah, differentiating 1/(sin y)^2 gives -2 cos y/(sin y)^3, not 1/cos y

:(
Click to expand...

Well I know I was wrong, it's a fairly straightforward integration IIRC. Have I got the method you're after though?

Edit: 1 / cos (y) = sec (y), and there's a formula for integration of sec (y). I DON'T KNOW IT THOUGH YO

Edit 2: I FOUND IT YO:

"squiggly line" (sec x dx ) = ln |sec x + tan x| + C
 
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Cool, assuming you're right that gives

ln |sec y + tan y| = Nx

sec y + tan y = e^{Nx}

...and how the beejesus do I find y in terms of x?

Edit: good call on the method, I was trying a load of stupid substitutions!

Edit 2: Score!

1/cos y + sin y/cos y = e^{Nx}

(1+sin y)/cos y = e^{Nx}

(1+sin y)/sqrt(1 - sin^2 y) = e^{Nx}

sqrt{ (1+sin y)/(1-sin y) } = e^{Nx}

(1+sin y)/(1-sin y) = e^{2Nx}

sin y (e^{2Nx} + 1) = e^{2Nx} - 1

y = arcsin [ (e^{2Nx} - 1) / (e^{2Nx} + 1 ]

y = arcsin (tanh Nx)

Cheers Carzy :)
 
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Arcade Fire said:
Cool, assuming you're right that gives

ln |sec y + tan y| = Nx

sec y + tan y = e^{Nx}

...and how the beejesus do I find y in terms of x?

Edit: good call on the method, I was trying a load of stupid substitutions!
Click to expand...

You'll have to do some messed up inverting or something. I can't remember; I can't even remember how to make x the subject of sin x = y :p Unless it's inverse (sin y) = x

And then you'll have to find the inverse of sec and tan. Which I haven't done in like two years, if at all.
 
It's okay, I already sorted it. Actually my method of substitution works out fine, as long as you pick the substitution y = arcsin z. It then reduces to something much simpler, which you can solve using partial fractions. Yes yes.
 
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