Just feeling stupid here...
y = x e^(dy/dx)
dy/dx = dx/dx.e^(dy/dx) + x. d^2y/dx^2.e^(dy/dx)
dy/dx = ( 1 + d^2y/dx^2) . e^(dy/dx)
Getting rid of the exponential...
y = x. dy/dx / ( 1 + d^2y/dx^2 )
Is this solvable? I don't know...
Anyway, for help on solving weird integrals, you can always refer to the old Abramowitz: http://people.math.sfu.ca/~cbm/aands/abr...stegun.pdf
Alternatively, if you're really really really lazy, there's wolfram alpha.
It says that you have a version of d'Alembert's equation in your hands.
y = x e^(dy/dx)
dy/dx = dx/dx.e^(dy/dx) + x. d^2y/dx^2.e^(dy/dx)
dy/dx = ( 1 + d^2y/dx^2) . e^(dy/dx)
Getting rid of the exponential...
y = x. dy/dx / ( 1 + d^2y/dx^2 )
Is this solvable? I don't know...
Anyway, for help on solving weird integrals, you can always refer to the old Abramowitz: http://people.math.sfu.ca/~cbm/aands/abr...stegun.pdf
Alternatively, if you're really really really lazy, there's wolfram alpha.
It says that you have a version of d'Alembert's equation in your hands.
