Non-linear ODEs

Came across this in my applied math grad course. The correct answer is incredibly simple, the equation of a circle. But in order to arrive at that solution a fair bit of geometry, algebra, and solution of a non-linear ODE is required.

A curve passing through (1,2) has the property that the length of a line drawn between the origin and intersecting perpendicular with the normal extending from any point along the curve is always numerically equal to the ordinate of that point on the curve. Find the equation of the curve.

Put another way: “Any curvature has a normal and a tangent. Draw the normal out and draw a line passing from the origin that is perpendicular to that (this line will be parallel to the tangent). The length of this line is equal to the ‘y’ of the point from which the normal emanates.”

Here’s a crude diagram:

Solution strategy is as follows:

We have some function y=f(x). We pick any point on f, and call that point (x,y) and draw a normal to the function at that point. We then draw a line from the origin such that it intersects with the normal perpendicularly at a point (a,b). The length of the vector (a,b) is equal to the ordinate of the point on the function from which the normal emanates, ‘y’. So that |(a,b)| = y.

What function satisfies this condition, and also passes through the point (1,2)?

Here’s the completed solution: BLAM!

Note the necessity of solving a non-linear (!) first order ODE to get the function! This is a special case ODE that has a ready analytic solution and is referred to as a Bernoulli ODE. It takes the general form dy/dx + P(x)y = Q(x)*(y^n).
Here’s a plot of the solution function:

Source: Schaum’s Advanced Mathematics for Engineers and Scientists, Ch2 Prob 83

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