Complex Trigonometry Question

[redtree]

Suppose y = f(x) * sin (kx), where k = wavenumber.

If \int y*dy = 3*kx, solve for f(x)

[tiny-tim]

Hi redtree! :smile:
Suppose y = f(x) * sin (kx), where k = wavenumber.

If \int y*dy = 3*kx, solve for f(x)

erm … \int y*dy\ =\ \frac{1}{2}\,y^2 :confused:

do you mean \int_0^x y(z)*dz\ =\ 3\,kx ?

If so, just differentatiate both sides. :smile:

[HallsofIvy]

If you really do mean \int y dy= 3kx, then (1/2)y2+ C= 3kx so
y= f(x)sin(kx)= \sqrt{6kx- 2C}
and
f(x)= \frac{\sqrt{6kx- 2C}}{sin(kx)}
where C and be any constant.

But I suspect tiny-tim is right.