Differential equations problem

In summary, a curve with Cartesian equation y=f(x) passes through the origin and divides every rectangle formed by lines drawn parallel to the coordinate axes into two regions A and B, with one region having an area equal to n times the other. The function f(x) that satisfies this condition is f(x)=cx, where c is a real number. Additionally, another possible solution is y=const. x^(1-k).
  • #1
5
0

Homework Statement


A curve with Cartesian equation y=f(x) passes through the origin. Lines drawn parallel to the coordiante axes through and arbitrary poing of the curve form a rectangle with two sides on the axes. The curve divides every such rectangle into two regions A and B, one of which has and area equal to n times the other. Find all such functions f.


Homework Equations





The Attempt at a Solution



Obviously f(x)=cx where c is a real number works, but there must be others. Any ideas?
 
Physics news on Phys.org
  • #2
[tex]\int[/tex]ydx (from 0 to x) = kxy ( for k = +- 1/n+1 or n/n+1).
This gives y = k(x y' + y) ; hence y =const. x^(1-k)
 
  • #3
The area below the curve is, of course, [itex]\int_0^{x} f(t)dt[/itex] while the area above the curve is [itex]\int_0^{x} f(x)- f(t) dt= xf(x)- \int_0^{x} f(t) dt[/itex] Saying one is n times the other means that [itex]n\int_0^{x}f(t)dt= xf(x)- \int_0^x f(t)dt[/itex] so [itex]xf(x)= (n+1)\int_0^x f(t) dt[/itex].

Differentiating both sides of that with respect to x will give you a differential equation for f(x)- xdf/dx+ f(x)= (n+1)f(x) so that xdf/dx= n f(x). That is separable and easily integrable.
 

Suggested for: Differential equations problem

Back
Top