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help solving this differential equation

Hi all, i'm trying to solve the following DE which I have attached as an image, as well as written it in the title box as my Latex sucks.

$\displaystyle 2sinh(x)y'' - 2cosh(x)y' + ysinh^3(x) = 0$

Attachment 30241

If someone can explain the method used to solve this, that would be really helpful :)

Thanks

Re: 2sinh(x)y'' - 2cosh(x)y' + ysinh^3(x) = 0

Here's what I might try

First $\dfrac{d}{dx} \left(\dfrac{y'}{\sinh x}\right) = \dfrac{\sinh x \,y '' - \cosh x \,y'}{\sinh ^2 x}$

So your DE becomes

$2 \dfrac{d}{dx} \left(\dfrac{y'}{\sinh x}\right) +\sinh x \,y = 0$

or

$2 \dfrac{1}{\sinh x } \,\dfrac{d}{dx} \left(\dfrac{y'}{\sinh x}\right) +y = 0$

Now try a new substitution. Let $ t = \cosh x$.

See how that goes.