Also O.D.E stuck on integration step

  • Thread starter asdf1
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  • #1
asdf1
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In the problem
y` (cosh^2)x - (sin^2)y = 0, y(0)=pi/2
i'm also stuck on the integration step...
my first step is
dy/((sin^2)y)=dx/(cosh^2)x
however, i don't how to integrate the right side~
 
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Answers and Replies

  • #2
TD
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You could try using the definition:

[tex]\cosh \left( x \right) \equiv \frac{{e^x + e^{ - x} }}{2}[/tex]
 
  • #3
asdf1
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ok, i'll try that~
thanks!
 
  • #4
lurflurf
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Formula for hyperbolic functions closely mirror those for circular functions
tan'(x)=(sec(x))^2
tanh'(x)=(sech(x))^2
Osbornes Rule is a way to do this in general
http://mathworld.wolfram.com/OsbornesRule.html
 
  • #5
asdf1
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wow~ i thought that only cos and sin have that kinda identity...
thanks! :)
 
  • #6
asdf1
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hmm...
if i use [tex]\cosh \left( x \right) \equiv \frac{{e^x + e^{ - x} }}{2}[/tex] to integrate cosh^2x=(e^2x + e^(-2x) +1)/2,
then [integrated(cosh)dx]=[(1/4)e^2x+(1/4)e^(-2x) + x/2]
but does that equal tanhx?
 
  • #7
Fermat
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I'm afraid not. It's the integral of sech²x that is equal to tanhx.

let y = tanhx
then
y = sinhx/coshx
y' = {coshx.(sinhx)' - sinhx.(coshx)'} / cosh²x
y' = {cosh²x - sinh²x} / cosh²x
y' = 1 / cosh²x
y' = sech²x
 
  • #8
asdf1
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ok, thanks!
 

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