During my study of calculus of variations I came across a nonlinear DE. I know that its solution is hyperbolic cosine with some constants yet do not know how to proceed since the function of interest is inside a square root. Also I am unfamiliar with the procedure of solving nonlinear differential equations. The equations is as follows: $$\frac{dy}{dx}=\pm\sqrt{\frac{\sigma g y + \lambda}{C} - 1} $$

EDIT: This equation is got after we try to minimize the potential energy of a cable hung between two poles with a constraint on its length a uniform linear mass density of $\sigma$ and $\lambda$ is a Lagrange multiplier.


Are you sure the solution is a combination of hyperbolic cosines?

Let's see. Re-write the ODE as follows:

$$ \pm \frac{dy}{\sqrt{k y - \beta}} = dx,$$

where $k = \sigma g /C$ and $\beta = 1-\lambda/C$. The equation above is equivalent to:

$$ \pm \frac{dy}{\beta^2 \sqrt{\frac{ky}{\beta} - 1}} = dx,$$

define $z:= ky/\beta$, so $dy = \beta dz/k $ and the differential equation becomes:

$$ \pm \frac{dz}{k \beta \sqrt{z-1}} = dx,$$

so a first integration gives us:

$$ \pm \frac{2 \sqrt{z-1}}{k \beta} = x + A,$$

being $A$ a constant of integration. Now you can explicitly solve for $z = z(x)$ as follows:

$$z(x) = 1+ \left( \frac{k\beta}{2} \right)^2 \left(x+A\right)^2.$$

You can obtain the solution for $y(x)$ in terms of the change of variable carried out before.


  • $\begingroup$ This agrees with the Wolfram Alpha solution, so I would say OP is mistaken (or make a mistake in typing the ODE). $\endgroup$ – Cameron Williams Jan 15 '14 at 0:10
  • $\begingroup$ I think that for the solution to be a combination of $cosh$ or $sinh$, the ODE's RHS should be $\sqrt{ky^2-\beta}$ instead. $\endgroup$ – Dmoreno Jan 15 '14 at 0:12
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    $\begingroup$ $cosh$ arises in catenaries, and the differential equation for a catenary is $\pm\frac{dy}{\sqrt{(y\!-\!\lambda)^{2}\!-\!a^{2}}}\;=\;\frac{dx}{a}$. The OP might want to check out this planet math article equation of catenary via calculus of variations for a full derivation. $\endgroup$ – 4ae1e1 Jan 15 '14 at 0:47
  • $\begingroup$ @Vesnog: in general, it's not a bad idea if when you suspect you know the functional form of the solution (e.g. $\cosh$, $\sin$, $\exp$ or whatever) to plug in a general form of it (e.g. $y=a\cosh(bx+c)$) and give it a shot. Sometimes an observation of the physical system could give a hint as to the functional form of the solution. If you're lucky, once you've plugged in the solution guess you'll discover what choice of constants $a, b, c$ would satisfy the equation and boundary condition(s). $\endgroup$ – Paul Safier Jan 15 '14 at 1:30
  • $\begingroup$ @Dmoreno I am not exactly sure but our instructor proposed that the solution was hyperbolic cosine with two arbitrary constants. Furthermore, I saw on Wolfram functions site that all 6 hyperbolic functions satisfy first order nonlinear DEs which suggested that this should be so in my case after a change of variables. $\endgroup$ – Vesnog Jan 15 '14 at 10:10

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