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Well $$y'''ctg(y')=-2(y'')^2$$ with the boundry conditions $$y(0)=1 \;;\;\; y'(0)=0 \;;\;\; y''(0)=1$$

As no $x$ is present in the initial DE it's safe to lower the order.

$$y'=z(y)$$ and $$y''=z'y'=z'z$$ $$y'''=z''z^2+z'^2z$$

And we get $$(z''z^2+z'^2z)ctg(z)+2z'^2z^2=0$$ now my problem, what do I do with the $ctg(z)$ ? How do I make this slightly nicer?

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    What is $ctg(x)$ ? – 2017-02-12
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    @Claude Liebovici $ctg(x)$=$cot(x)$=$\frac{cos(x)}{sin(x)}$ – 2017-02-13

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$$y'''ctg(y')=-2(y'')^2\\ \frac{y'''}{-2y''}=\frac{y''}{ctg(y')}\\ -\frac{1}{2}\ln y''+C=\int_0^{y'} \frac{\mathrm{d}t}{ctg(t)}=f(y')\\ C y''^{-1/2}=e^{f(y')}\\ C y''^{-1}=e^{2f(y')}\\ C =e^{2f(y')}y''\Leftarrow C=1 \text{ by initial condition}\\ x+A =\int^{y'} e^{2f(t)}\,\mathrm{d}t=g(y')\\ g^{-1}(x+A)=y'\Leftarrow A=g(0)\\ y=1+\int_0^xg^{-1}(x+g(0))\,\mathrm{d}t $$

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    A little tidying up to do and once I check if there are no problems you get the answer. – 2017-02-12