# Please suggest a substitution for solving or any method of

Please suggest a substitution for solving or any method of solving:
$$\displaystyle{y}{''}{\left({x}\right)}{\cot{{\left({y}{\left({x}\right)}\right)}}}={y}'{\left({x}\right)}^{{2}}+{c}$$

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esfloravaou
Substituting $$\displaystyle{u}={\sin{{\left({y}\right)}}}$$ one gets
$$\displaystyle{u}{''}={\left({\sin{{\left({y}\right)}}}\right)}{''}={\left({\cos{{\left({y}\right)}}}{y}'\right)}'={\cos{{\left({y}\right)}}}{y}{''}-{\sin{{\left({y}\right)}}}{y}'^{{2}}={c}{\sin{{\left({y}\right)}}}={c}{u}$$
which is, depending on the sign of c, an oscillation equation or an exponential function which is easily solvable for u and thus for y.
$u=\begin{cases}a_1\cos(\sqrt{|c|}x)+a_2\sin(\sqrt{|c|}x)&for\ c<0\\b_1+b_2x&for\ c=0\\c_1e^{\sqrt{c}x}+c_2e^{\sqrt{c}x}&for\ c>0\end{cases}$
###### Not exactly what youâ€™re looking for?
Serita Dewitt
$$\displaystyle{\cot{{y}}}{y}{''}={y}'^{{2}}+{c}$$
Substitute $$\displaystyle{y}'={p}\Rightarrow{y}{''}={p}{\frac{{{d}{p}}}{{{\left.{d}{y}\right.}}}}$$
$$\displaystyle{p}{p}'{\cot{{y}}}={p}^{{2}}+{c}$$
$$\displaystyle\int{\frac{{{2}{p}}}{{{p}^{{2}}+{c}}}}{d}{p}={2}\int{\tan{{\left({y}\right)}}}{\left.{d}{y}\right.}$$
$$\displaystyle{\ln}{\left|{p}^{{2}}+{c}\right|}=-{2}{\ln}{\left|{\cos{{\left({y}\right)}}}\right|}+{K}_{{1}}$$
$$\displaystyle{p}^{{2}}={\frac{{{K}_{{1}}}}{{{{\cos}^{{2}}{\left({y}\right)}}}}}-{c}$$
$$\displaystyle{y}'=\sqrt{{{\frac{{{K}_{{1}}}}{{{{\cos}^{{2}}{\left({y}\right)}}}}}-{c}}}$$
$$\displaystyle\int{\frac{{{\left.{d}{y}\right.}}}{{\sqrt{{{\frac{{{K}_{{1}}}}{{{{\cos}^{{2}}{\left({y}\right)}}}}}-{c}}}}}}={x}+{K}_{{2}}$$
with $$\displaystyle{u}={\sin{{\left({y}\right)}}}$$
$$\displaystyle\int{\frac{{{d}{u}}}{{\sqrt{{{K}_{{1}}+{c}{u}^{{2}}}}}}}={x}+{K}_{{2}}$$