Question

Find a curve y=y(x) through (0, 2) such that the normal to the curve at any point (xo, y(xo)) intersects the x axis at xI=xo+1.

Answer 1

@SithsAndGiggles

Answer 2

Same approach as before to determine the setup - We find the equation of a tangent line to \((x_0,y_0)\):
\[y-y_0=y'(x_0)(x-x_0)\]
However, this time we're concerned with the normal line to the curve at \((x_0,y_0)\). This line is perpendicular to the tangent line at the same point, which means it has a negative reciprocal slope:
\[y-y_0=-\frac{1}{y'(x_0)}(x-x_0)\]
We want this line to pass through the x-axis when \(x=x_1=x_0+1\), i.e. the point \((x_0+1,\,0)\).

Plugging this point into the equation of the normal line, we get
\[0-y_0=-\frac{1}{y'(x_0)}(x_0+1-x_0)\]
which, upon rewriting, give the separable ODE
\[y_0=\frac{1}{{y_0}'}~~\iff~~y_0\,dy_0=dx_0\]
(Note: \(y_0=y(x_0)\) and \({y_0}'=y'(x_0)\).)

Answer 3

Thank you!