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I'm trying to solve this equation but at the end I'm stuck and can't reache the answer. I use the substitutions $y'=p$ and $y'' = p'\cdot p$: $$y \cdot p'p - p^2 - 1 = 0 \implies y\cdot p \frac {dp}{dy} - (p^2 + 1) = 0 \implies \int \frac {p}{p^2 + 1}dp = \int \frac {dy}{y} \implies \frac{1}{2} \ln \left|C_1(p^2 + 1) \right | = \ln y \implies y = C_1 \sqrt{p^2+1} \implies p = \sqrt {C_1y^2 - 1} $$

Next step: $$ y' = \sqrt {C_1 y^2 - 1} \implies \frac{dy}{dx} = \sqrt {C_1 y^2 - 1} \implies \int \frac {dy}{C_1 \sqrt{ y^2 - \frac {1}{C_1^2} }} = \int dx \implies \\ C_1 \ln \left |y + \sqrt{y^2 - C^2_1} \right | = x + C_2 \implies y + \sqrt{y^2 - C_1^2} = \exp {\frac {x+C_2}{C_1}} $$

Here I don't know how to go on. The answer should be $ y = \frac {C_1}{2} \left ( \exp(\frac{x+C_2}{c_1}) + \exp(-\frac{x+C_2}{c_1}) \right )$

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  • $\begingroup$ One apparent solution is $y=\cosh(x-a)$. $\endgroup$ – alex.jordan Jun 20 '15 at 11:40
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    $\begingroup$ If you substitute $y'=p$, why would that be consistent with $y''=p'\cdot p$? It wouldn't, unless $p$ is $1$. $\endgroup$ – alex.jordan Jun 20 '15 at 11:42
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    $\begingroup$ @alex.jordan The OP is referring to the trick of searching for solutions $y$ such that $y'=p(y)$, that is, for every $x$, $$y'(x)=p(y(x)).$$ $\endgroup$ – Did Jun 21 '15 at 9:03
  • $\begingroup$ $\frac{(y^{\prime})^2+1}{y^2}=C$ is a first integral to the ODE $yy^{\prime\prime} - (y^{\prime})^2 = 1$. $\endgroup$ – Qmechanic Jun 21 '15 at 12:10
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    $\begingroup$ Oh, $p$ depends on $y$, I see. $\endgroup$ – alex.jordan Jun 21 '15 at 12:41
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Maybe your $C_1$ and $C_2$ are different to the answer's. But let solve your equation: $$ y+\sqrt{y^2-C_1^2}=\exp(\frac{x+C_2}{C_1})\\ \sqrt{y^2-C_1^2}=\exp(\frac{x+C_2}{C_1})-y\\ y^2-C_1^2=\exp(2\frac{x+C_2}{C_1})-2y\exp(\frac{x+C_2}{C_1})+y^2\\ y=\frac{1}{2}(\exp(\frac{x+C_2}{C_1})+C_1^2\exp(-\frac{x+C_2}{C_1})) $$

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  • $\begingroup$ Your solution doesn't match the answer in my textbook (there isn't $C^2_1$ in the second part of expression), but I obtain the same results. So I think there is a mistake in the textbook. $\endgroup$ – flipback Jun 24 '15 at 3:24
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Divide by $y^2$. Then you have $(y'/y)' = 1/y{^2}$. I'm not sure if this helps. The other way would be to take the original equation and differentiate. You get $y y''' = y'y'' $. Rearranging, we have $y'/y = y'''/y''$. This gives $ln (y'') = ln (y) + c$, which gives $y'' = e^{c} y $. Now with $u^2 = e^{c}$, $y=a cosh (ut) + b sinh (ut) $ is a general solution. Also confirms @alex.jordan 's guess. Please excuse my fillers in between, I will learn laTex soon.

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  • $\begingroup$ The right hand side should be $1/y^2$ rather than $(-1/y)'$ $\endgroup$ – Lion Jun 22 '15 at 5:45
  • $\begingroup$ You are right. Correcting it. $\endgroup$ – Shailesh Jun 22 '15 at 6:12
  • $\begingroup$ @Lion. I have also added a solution. Thanks for pointing out the error $\endgroup$ – Shailesh Jun 22 '15 at 6:38
  • $\begingroup$ It seems you've given the best clew. But I can't work out the passage from $y'' = e^c y $ to $y = acosh(ut) + bsinh(ut)$. Sorry, my maths is very limited to understand that kind of tricks. $\endgroup$ – flipback Jun 22 '15 at 15:48
  • $\begingroup$ @flipback Check out how the functions $cosh$ and $sinh$ are defined. That will give you some clue. $\endgroup$ – Shailesh Jun 22 '15 at 15:59
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You obtained : $$y + \sqrt{y^2 - \frac{1}{C_1^2}} = \exp {\frac {x+C_2}{C_1}} $$ it remains to solve this equation for $y$ $$\sqrt{y^2 - \frac{1}{C_1^2}} = -y+\exp {\frac {x+C_2}{C_1}} $$ $$y^2 - \frac{1}{C_1^2} = \left( -y+\exp {\frac {x+C_2}{C_1}}\right)^2 $$ Then, simplify and express $y$.

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