Why can both sides of 3X'/X = -2Y'/Y be set equal to a constant?

One side depends only on x and the other only on y

PDE by separation of variables

Short answer

Using separation u=XY, we get 3X'/X = -2Y'/Y = constant k, leading to u(x,y) = C e^(k(x - 3y/2)) = C e^(k(2x-3y)/2) with arbitrary constants C and k.

Find the solution of the equation 3uₓ + 2u_y = 0 by the method of separation of the variables.

Let u = X(x)Y(y). Then uₓ = X′Y, u_y = XY′, and 3uₓ + 2u_y = 0 becomes

3X′Y + 2XY′ = 0 → 3X′/X = −2Y′/Y = 3k (a constant)

Answer: u(x, y) = C exp[ k(x − 3y/2) ] = C e^(k(2x − 3y)/2), with C and k arbitrary constants.

Check: uₓ = ku and u_y = −(3k/2)u, so 3uₓ + 2u_y = 3ku − 3ku = 0.

✓ Answer checked by Claude Opus 5.5 (high effort)