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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.

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.