difficulty in manipulaton (eliminating term "sinh y"??)

[Rohit93]

I am having difficulty in this manipulation:

Illustration 6:

. . . . .Solve \(\displaystyle \dfrac{dy}{dx}\, =\, \dfrac{x \, \sin(x)}{2\, e^y\, \sinh(y)}\)

Solution:

. . . . .Rewrite the equation as:

. . . . . . . . . .. .\(\displaystyle x\, \sin(x)\, dx\, =\, 2\, e^y\, \sinh(y)\, dy\)

. . . . . . . . . .\(\displaystyle \int\, x\, \sin(x)\, dx\, =\, 2\, \int\, e^y\, \sinh(y)\, dy\, +\, c\)

. . . . . . . . . . . . . . . . . . ..\(\displaystyle =\, 2\, \int\, e^y\, \dfrac{\left(e^y\, -\, e^{-y}\right)}{2}\, dy\, +\, c\)

. . . . . . . . . . . . . . . . . . ..\(\displaystyle =\, \int\, \left(e^{2y}\, -\, 1\right)\, dy\, +\, c\)

This is a differential equation problem, my doubt here is i am not able to understand how did they eliminate the term 'sinh y' .
Is there some sort of formula that I am missing????? please help thanks in advance

 

[JumboJam]

I think that is just a trig identity--by definition, sinh(x) = (e^x - e^-x) / 2

 

[Deleted member 4993]

I think that is just a trig identity--by definition, sinh(x) = (e^x - e^-x) / 2

Correct.

Similarly:

cosh(Θ) = (e^Θ + e^-Θ)/2

and

cos(Θ) = (e^iΘ + e^-iΘ)/2

sin(Θ) = (e^iΘ - e^-iΘ)/(2i)

 

[Rohit93]

Thanks i knew i was missing something.

 

[HallsofIvy]

I think that is just a trig identity--by definition, sinh(x) = (e^x - e^-x) / 2

True, though I wouldn't call it a "trig" identity! Rather a "hyperbolic" identity.