Consider R=((1+y'^2)^(3/2))/(|y''|), N = y√(1+y'^2), R=λ N+μ…

Consider $R=\frac{(1+y'^2)^{3/2}}{|y''|}, \qquad N = y\sqrt{1+y'^2}, \qquad R=\lambda N+\mu$. 1° Find the curves for which the radius of curvature equals twice the length of the normal, $R=2N$. 2° More generally, find the curves satisfying the linear relation $R=\lambda N+\mu$ between the radius of curvature and the normal.

Step 1: Preliminaries

For a curve $y=y(x)$, the length of the normal segment (from the point of tangency to the $x$-axis, along the normal) is $N=y\sqrt{1+y'^2}$, and the radius of curvature is $R=(1+y'^2)^{3/2}/|y''|$. Assume $y''>0$ throughout, so $|y''|=y''$.

Step 2: Part 1°: the condition $R=2N$

Setting $R=2N$ gives

\[ \frac{(1+y'^2)^{3/2}}{y''}=2y\sqrt{1+y'^2} \quad\Longrightarrow\quad 1+y'^2 = 2y\,y''. \]

Let $p=y'$, regarded as a function of $y$, so that $y''=p\,\dfrac{dp}{dy}$; the equation becomes

\[ 1+p^2 = 2yp\,\frac{dp}{dy} \quad\Longrightarrow\quad \frac{2p\,dp}{1+p^2}=\frac{dy}{y}. \]

Integrating, $\ln(1+p^2)=\ln y + \mathrm{const}$, i.e. $1+p^2=C_1^2y$, so $p^2=C_1^2y-1$. Separating variables once more, $\frac{dy}{\sqrt{C_1^2y-1}}=\pm\,dx$.

With $u=C_1^2y-1$, $du=C_1^2\,dy$, the left side integrates to $\dfrac{2\sqrt{C_1^2y-1}}{C_1^2}$, so

\[ \frac{2\sqrt{C_1^2y-1}}{C_1^2} = \pm(x-C_2). \]

Squaring and solving for $y$, and renaming the constants as $A=C_1^2/4$, $B=C_2$, yields the family of parabolas

\[y = A(x-B)^2 + \frac{1}{4A}\]

(Independently verified: for this $y$, the ratio $R/N$ simplifies identically to $2$ for all $x$, regardless of $A$ and $B$.)

Step 3: Part 2°: the general relation $R=\lambda N$

Consider first $\mu=0$, i.e. $R=\lambda N$. The same computation as above, with $2$ replaced by $\lambda$, gives $1+y'^2=\lambda y\,y''$, and with $p=y'$, $y''=p\,dp/dy$, \[1+p^2 = \lambda y p\,\frac{dp}{dy} \quad\Longrightarrow\quad \frac{p\,dp}{1+p^2}=\frac{dy}{\lambda y}.\]

Integrating, $\tfrac12\ln(1+p^2)=\tfrac1\lambda\ln y+\mathrm{const}$, i.e.

\[ 1+p^2 = C^2y^{2/\lambda}, \]

so that $p^2=C^2y^{2/\lambda}-1$ and

\[ \frac{dy}{\sqrt{C^2y^{2/\lambda}-1}} = \pm\,dx, \]

an integral that is elementary whenever $2/\lambda$ is a simple rational exponent — in particular for $\lambda=1$ and $\lambda=2$.

Step 4: The case $\mu\neq0$

When the relation is $R=\lambda N+\mu$ with $\mu\neq0$, the same substitution $p=y'$, $y''=p\,dp/dy$ still reduces the second-order equation to a first-order equation for $p(y)$:

\[ 1+p^2 = y p\,\frac{dp}{dy}\Bigl(\lambda + \frac{\mu}{y\sqrt{1+p^2}}\Bigr), \]

which is no longer separable in closed elementary form in general, but remains a well-posed first-order ODE in $p(y)$, solvable at least by quadrature (numerically or via series, depending on $\lambda,\mu$).

\[\begin{aligned} \lambda=1:&\quad y=\tfrac1C\cosh\bigl(C(x-C_2)\bigr) &&\text{(catenaries)}\\[4pt] \lambda=2:&\quad y=A(x-B)^2+\tfrac1{4A} &&\text{(parabolas)}\\[4pt] \text{general }\lambda:&\quad \displaystyle\int\frac{dy}{\sqrt{C^2y^{2/\lambda}-1}}=\pm x+C_2 \end{aligned}\]