Find the family of curves for which at every point the sum…
Find the family of curves for which at every point the sum of the length of the normal and the subnormal equals unity. Find the curve passing through $(0,0)$.
Explanation of Method
Express the normal length $y\sqrt{1+y'^2}$ and the subnormal $yy'$ in terms of $x,y,y'$, set their sum to $1$, and isolate $y'$ by squaring after isolating the radical.
Step 1: Set up the condition
$y\sqrt{1+(y')^2}+yy'=1$ (taking $y,y'>0$). Isolate the radical: $y\sqrt{1+(y')^2}=1-yy'$.
Step 2: Square and simplify
$y^2[1+(y')^2]=1-2yy'+y^2(y')^2$. The $y^2(y')^2$ terms cancel: $y^2=1-2yy'$, so
\[y'=\frac{1-y^2}{2y}.\]
Step 3: Separate and integrate
$\dfrac{2y}{1-y^2}dy=dx$. With $u=1-y^2$: $-\ln|1-y^2|=x+C_1$, so $1-y^2=Ce^{-x}$.
Step 4: Apply the initial condition
At $(0,0)$: $1-0=Ce^0\Rightarrow C=1$.
Result
Final answer:
General family: \[y^2=1-Ce^{-x},\qquad C\in\mathbb{R}\]
Particular curve through $(0,0)$ (for $x\ge0$, approaching the asymptote $y=1$):
\[\boxed{\,y=\sqrt{1-e^{-x}}.}\]