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Solve a partial differential equation

Short answer

z = (1+cos x) cos y. Integrating ∂²z/∂x∂y gives ∂z/∂y = −(1+cos x) sin y; condition yields no extra term, then integration in y gives solution.

Solve the partial differential equation ∂²z/∂x∂y = sin x sin y given that ∂z/∂y = −2 sin y when x = 0, and z = 0 when y = π/2.

Integrate the equation twice, once with respect to each variable, and use the conditions to find the arbitrary functions.

∂²z/∂x∂y = sin x sin y
Integrate w.r.t. x:   ∂z/∂y = −cos x sin y + g(y)
Use x = 0, ∂z/∂y = −2 sin y:
                      −2 sin y = −sin y + g(y)   so  g(y) = −sin y
Hence                 ∂z/∂y = −(1 + cos x) sin y
Integrate w.r.t. y:   z = (1 + cos x) cos y + h(x)
Use y = π/2, z = 0:   0 = 0 + h(x)   so  h(x) = 0

Answer: z = (1 + cos x) cos y

Check: ∂z/∂x = −sin x cos y, so ∂²z/∂x∂y = sin x sin y.

✓ Answer checked by Claude Opus 5.5 (high effort)

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