Integrating 1/cosh²x in separable ODE

[asdf1]

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~

 

[TD]

You could try using the definition:

$$\cosh \left( x \right) \equiv \frac{{e^x + e^{ - x} }}{2}$$

 

[asdf1]

ok, i'll try that~
thanks!

 

[lurflurf]

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

 

[asdf1]

wow~ i thought that only cos and sin have that kinda identity...
thanks! :)

 

[asdf1]

hmm...
if i use $$\cosh \left( x \right) \equiv \frac{{e^x + e^{ - x} }}{2}$$ 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?

 

[Fermat]

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

 

[asdf1]

ok, thanks!