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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^{2}y^{2}+f(y)\,

Integrate wrt the partial y:

z(x,y)={\frac  {1}{12}}x^{2}y^{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:



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^{2}y^{3}+\sin(y)-1+e^{x}\,

Partial Differential Equations

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