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## Source

IM Gelfand, SV Fomin. 1963. *Calculus of Variations*. Mineola, NY: Dover. Tr. RA Silverman. p. 20.

## Problem

At one point in Sec. 4.2, Example 2, Gelfand and Fomin move from

\( x + C_1 = C \ln \frac{y + \sqrt{y^2 – C^2}}{C} \)

to

\( y = C \cosh \frac{x + C_1}{C} \)

This move did not seem immediately transparent to me, so I fill in the details here.

## Walkthrough

When I started this, all I remembered about \( \cosh \) is that it is expressed as a function of some exponential terms. I realized that in isolating \( y \)—something I would have to do anyway—I would end up with exponential terms (see steps 3 and 4), so my strategy was just to isolate \( y \) and then look up \( \cosh \) to see whether it was algebraically equivalent to what I had at that point.

We start with

\( x + C_1 = C \ln \frac{y + \sqrt{y^2 – C^2}}{C} \tag{1} \)

Divide through by \( C \):

\( \frac{x + C_1}{C} = \ln \frac{y + \sqrt{y^2 – C^2}}{C} \tag{2} \)

Since for all expressions \( \psi \) and \( \phi \), \( \psi = \phi \) implies \( e^\psi = e^\phi \), we can change 2 to

\( e^{\frac{x + C_1}{C}} = e^{\ln \frac{y + \sqrt{y^2 – C^2}}{C}} \tag{3} \)

Since for any expression \( \psi \), \( e^{\ln \psi} = \psi \), we get

\( e^{\frac{x + C_1}{C}} = \frac{y + \sqrt{y^2 – C^2}}{C}. \tag{4} \)

Multiply through by \( C \):

\( C e^{\frac{x + C_1}{C}} = y + \sqrt{y^2 – C^2} \tag{5} \)

Subtract \( y \) from both sides:

\( C e^{\frac{x + C_1}{C}} – y = \sqrt{y^2 – C^2} \tag{6} \)

Square both sides, remembering that for any expression \( \psi \), \( (e^\psi)^2 = e^{2\psi} \):

\( C^2e^{2\left(\frac{x+C_1}{C}\right)} -2Ce^{\frac{x+C_1}{C}}y + y^2 = y^2 – C^2 \tag{7} \)

Subtract \( y^2 \) from both sides:

\( C^2e^{2\left(\frac{x+C_1}{C}\right)} -2Ce^{\frac{x+C_1}{C}}y = – C^2 \tag{8} \)

Divide through by \( C \):

\( Ce^{2\left(\frac{x+C_1}{C}\right)} -2e^{\frac{x+C_1}{C}}y = – C \tag{9} \)

Add \(2e^{\frac{x+C_1}{C}}y + C \) to both sides:

\( Ce^{2\left(\frac{x+C_1}{C}\right)} + C = 2e^{\frac{x+C_1}{C}}y \tag{10} \)

Divide through by \( 2e^{\frac{x+C_1}{C}} \) and then reverse the order:

\( y = \frac{Ce^{2\left(\frac{x+C_1}{C}\right)} + C}{2e^{\frac{x+C_1}{C}}} \tag{11} \)

At this point, I look up \( \cosh \) and discover that by one of its definitions, \( \cosh \psi = \frac{e^{2\psi} + 1}{2e^\psi} \). Comparing this with 11, I see what to do. Pull \( C \) out of the numerator:

\( y = C \frac{e^{2\left(\frac{x+C_1}{C}\right)} + 1}{2e^{\frac{x+C_1}{C}}} \tag{12} \)

This has the form of \( \cosh \psi \) described above, with \( \psi = \frac{x+C_1}{C} \), so we have

\( y = C \cosh \left( \frac{x+C_1}{C}\right) \tag{13} \)

And there you have it.