economia: modelo de superficie de la ecuación diferencial de primer orden

f(x) = e^{(m/x^{2})}
ln( f(x) ) = (m/x^{2})


d_{y}[g(y)] + (-1)·( m/x^{3} )·g(y) = 0


g(y) = e^{(m/x^{3})·y}


(1/ln(e^{(m/x^{3})·(x^{2}/m)})) = x


(1/ln(e^{(m/x^{3})·( 1/ln( f(x) ) )})) = x


( 1/ln( g( 1/ln(f(x)) ) ) ) = x


m = 2 cajas & x = 20 cm ==> ( 1/ln( f(20) ) ) = 200€


d_{y}[g(y)] + (-1)·(1/4000)·g(y) = 0


( 1/ln( g(200) ) ) = 20€


m = 3 cajas & x = 60 cm ==> ( 1/ln( f(60) ) ) = 1200€


d_{y}[g(y)] + (-1)·(1/72000)·g(y) = 0


( 1/ln( g(1200) ) ) = 60€