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Please suggest a substitution for solving:

$$y'' \cot( y ) = (y'){^2} +c $$

Thanks in advance.

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closed as off-topic by Servaes, Jyrki Lahtonen, Dylan, Xander Henderson, José Carlos Santos Jul 27 '18 at 14:33

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  • $\begingroup$ Still trying to guess.. in the direction of log sec/ cosec $\phi,\phi^{'}$ etc. so far no luck. $\endgroup$ – Narasimham Jul 26 '18 at 8:06
  • $\begingroup$ is that $\cot(yy'')$ or $\cot(y)y''$? $\endgroup$ – LutzL Jul 26 '18 at 8:21
  • $\begingroup$ The latter; $ y^{''}\cot (y) $ $\endgroup$ – Narasimham Jul 26 '18 at 8:25
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$$\cot y \, y^{''} = y^{'}{^2} +c$$ Substitute $y'=p \implies y''=p\frac {dp}{dy}$ $$pp'\cot y = p^2 +c$$ $$\int \frac {2p}{ p^2 +c}dp=2\int \tan(y) dy$$ $$\ln|p^2 +c|=-2\ln |\cos(y)|+K_1$$ $$p^2 =\frac {K_1} {\cos^2(y)}-c$$ $$y'=\sqrt{\frac {K_1} {\cos^2(y)}-c}$$ $$\int \frac {dy}{\sqrt{\frac {K_1} {\cos^2(y)}-c}}=x+K_2$$ With $u=\sin(y)$ $$\int \frac {du}{\sqrt{K_1+cu^2}}=x+K_2$$

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Substituting $u=\sin(y)$ one gets $$ u''=(\sin(y))''=(\cos(y)y')'=\cos(y)y''-\sin(y)y'^2=c\sin(y)=cu $$ which is, depending on the sign of $c$, an oscillation equation or an exponential function which is easily solvable for $u$ and thus for $y$. $$ u=\begin{cases} a_1\cos(\sqrt{|c|}x)+a_2\sin(\sqrt{|c|}x)&\text{ for }c<0,\\ b_1+b_2x&\text{ for }c=0,\\ c_1e^{\sqrt{c}x}+c_2e^{\sqrt{c}x}&\text{ for }c>0. \end{cases} $$

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  • $\begingroup$ Pretty nice way around it ! Fancier than my brutal substitution. $\endgroup$ – Rebellos Jul 26 '18 at 8:30
  • $\begingroup$ The form reminded me of the more common second derivative of $\exp(y)$. $\endgroup$ – LutzL Jul 26 '18 at 8:31
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HINTS :

If $y$ is a function of $x$, namingly $y(x)$ and $c$ is just a constant such that $c \in \mathbb R$, then a (very complicated) solution can be yielded after a string of operations using the substitution $$v(y) = \frac{\mathrm{d}y(x)}{\mathrm{d}y}$$ which gives $$\frac{\mathrm{d}^2y(x)}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}\bigg(\frac{\mathrm{d}y(x)}{\mathrm{d}x}\bigg)=\frac{\mathrm{d}v(y)}{\mathrm{d}x}=\frac{\mathrm{d}v(y)}{\mathrm{d}y}\frac{\mathrm{d}y}{\mathrm{d}x}=v(y)\frac{\mathrm{d}v(y)}{\mathrm{d}x}$$ and thus integrating solving for $v(y)$ and substituting back to find an expression in terms only of $y(x)$ and $x$.


If $y$ is a function of $c$, then there isn't a solution in terms of standard functions. Specifically, the solution is

$$y(c) = \arcsin\big[c_1\big(c_2\mathrm{Bi}(c) + \pi \mathrm{Ai}(c)\big)\big]$$

which gives some rather interesting plots-properties while sampling an initial $y(0)$, where $\mathrm{Ai}(x)$ is the Airy function and $\mathrm{Bi}(x)$ is the Airy Bi function.

$\qquad\qquad\qquad\qquad\qquad$enter image description here

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