# PDE3

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

Since $z_{xy} = \frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right)=\frac{1}{2}xy^2\,$

Integrate wrt the partial x:

$\frac{\partial z}{\partial y} = \frac{1}{4}x^2y^2 + f(y)\,$

Integrate wrt the partial y:

$z(x,y) = \frac{1}{12}x^2y^3 + F(y) + g(x)\,$ where $\frac{\partial F}{\partial y} = f(y)\,$.

This is the general solution because it has two arbitrary and independent functions. Now take into account the IC's and solve for the most useful term:

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

$z(0,y) = F(y)+g(0)=\sin(y),\,\,F(y)=\sin(y)-g(0)=\sin(y)-1+F(0)\,$

Plug the function values back into the original PDE to get the unique solution for this set of IC's.

$z(x,y) = \frac{1}{12}x^2y^3+\sin(y)-1+e^x\,$

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