Solve (y)/(√(1+y'^2))=1 using the parameter method

Solve $\dfrac{y}{\sqrt{1+y'^2}}=1$ using the parameter method.

Step 1: Solve for $y$

$y=\sqrt{1+p^2}$.

Step 2: Compute $dx$ (checked symbolically)

\texttt{sympy} gives $\dfrac{dy}{dp}=\dfrac{p}{\sqrt{1+p^2}}$, so \[dx=\frac{dy}{p}=\frac{1}{\sqrt{1+p^2}}\,dp,\] matching the source.

Step 3: Integrate

\texttt{sympy.integrate}$\left(1/\sqrt{1+p^2},p\right)=\operatorname{asinh}(p)$, so \[x=\operatorname{asinh}(p)+C \iff p=\sinh(x-C).\]

Step 4: Eliminate $p$

$y=\sqrt{1+p^2}=\sqrt{1+\sinh^2(x-C)}=\sqrt{\cosh^2(x-C)}=\cosh(x-C)$ (since $\cosh>0$ always). This gives the catenary $y=\cosh(x-C)$.

Step 5: End-to-end verification

With $y=\cosh(x-C)$, $y'=\sinh(x-C)$, \texttt{sympy} simplifies \[\frac{y}{\sqrt{1+y'^2}}=\frac{\cosh(x-C)}{\sqrt{1+\sinh^2(x-C)}}\] to exactly $1$, confirming the original equation identically.