solution of (∂^2 z)/∂x∂y=x^2 y is ?

Question

It is related to partial differential equations

Answer 1

∂²z/∂x∂y=∂(∂z/∂y)/∂x, therefore ∂z/∂y=∫x²y∂x=x³y/3+f(y).

∂z/∂y=x³y/3+f(y), z=(x³/3)∫y∂y+∫f(y)dy=(x³/3)(y²/2)+∫f(y)dy+g(x).

∂²z/∂x∂y=∂(∂z/∂x)/∂y, therefore ∂z/∂x=∫x²y∂y=x²y²/2+h(x).

∂z/∂x=x²y²/2+h(x), z=(y²/2)∫x²∂x+∫h(x)dx=(y²/2)(x³/3)+∫h(x)dx+j(y).

∂(∂z/∂y)/∂x=∂(∂z/∂x)/∂y⇒(x³/3)(y²/2)+∫f(y)dy+g(x)=(y²/2)(x³/3)+∫h(x)dx+j(y).

Therefore:

j(y)=∫f(y)dy and g(x)=∫h(x)dx, hence z=x³y²/6+g(x)+j(y).

g(x) and j(y) are unknown functions.

To check if this is correct let's invent g(x) and j(y):

Let g(x)=x⁵+x³+2x+1, so h(x)=5x⁴+3x²+2, and j(y)=ln(y)+sin(y), so f(y)=1/y+cos(y).

z=x³y²/6+x⁵+x³+2x+1+ln(y)+sin(y).

∂z/∂x=x²y²/2+5x⁴+3x²+2, ∂z/∂y=x³y/3+1/y+cos(y),

∂(∂z/∂y)/∂x=x²y, ∂(∂z/∂x)/∂y=x²y so ∂²z/∂x∂y=x²y. The example shows that z=x³y²/6+g(x)+j(y), where the functions are arbitrary.