# PDE2

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

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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