PDE2

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$z_{xy} = x^2y, \,\,z(x,0)=x^2, \,z(1,y)=\cos(y)\,$

Write $\frac{\partial^2 z}{\partial x \partial y} = x^2y\,$ as

$\frac{\partial}{\partial x}\left( \frac{\partial z}{\partial y} \right) = x^2y\,$

Integrate wrt the partial $x$ to get:

$\frac{\partial z}{\partial y} = \frac{1}{3}x^3y + f(y)\,$

Integrate wrt the partial $y$ to get:

$z(x,y) = \frac{1}{6}x^3y^2+F(y)+g(x)\,$ where $\frac{\partial F}{\partial y} = f(y)\,$ .

Since the above formula for $z (x, y)$ has two arbitrary and independent functions, it is the general solution to the PDE. Now match this to each IC, and solve for the most helpful term:

$z(x,0) = F(0) + g(x) = x^2, \,\,g(x) = x^2-F(0)\,$

$z(1,y) = \frac{1}{6}y^2 + F(y) + g(1) = \cos(y), \,\,F(y) = \cos(y)-\frac{1}{6}y^2-1+F(0)\,$

Now plug these function values into the general solution to find the unique solution for this set of IC's.

$z(x,y) = \frac{1}{6}x^3y^2 + \cos(y)-\frac{1}{6}y^2-1+x^2\,$

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