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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.

by Top Rated User (1.2m points)

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