Let me begin with a problem coming from geometry and its solution. Here is the statement: Find curves y=y(x) in the Cartesian plane such that every point P(x0,y(x0)) on the curve is the midpoint of the line segment tangent to the curve at P and with endpoints on the coordinate axes. And here is the solution: The tangent line (to the curve) at the point (x0,y(x0)) is y(x)=y(x0)+y(x0)(xx0). Denote its y-intercept by yI. Recall that it is obtained by setting x=0 in the equation of the line. Then yI=y(x0)y(x0)x0 Also, yI=2y(x0 ) (by the assumption about the point of tangency in the statement of the problem) so (using this in *) 2y(x0)=y(x0)x0y(x0) Since this must be true for every point P on the curve, we drop the subscript to get the first order, linear differential equation y+xy=0. Its complete solution is yc(x)=K/x, for arbitrary real K. Thus the curves satisfying the condition in the statement of the problem are those hyperbolas. Problem 1 Imitate the solution to the problem above to find all the curves y(x) in the Cartesian plane's first quadrant that satisfy the following geometric condition: The part of the tangent line at (x,y(x)) between that point and the x-axis is bisected by the y-axis. (Note: If xI is the x-intercept of the tangent line, then the geometric condition can be translated as 2x+xI=0.)