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- Feb 24th 2014, 05:09 AM #1

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## help solving this differential equation

- Mar 3rd 2014, 03:19 PM #2
## 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.