- **Differential Equations**
(*http://mymathforum.com/differential-equations/*)

- - **How can I solve this PDE?**
(*http://mymathforum.com/differential-equations/339150-how-can-i-solve-pde.html*)

How can I solve this PDE?-(∂²u/∂x² + ∂²u/∂y²) + uk² = xk² zero boundary conditions on the boundary of the Square Domain (0,1)X(0,1). k^2 is a constant |

All times are GMT -8. The time now is 05:52 PM. |

Copyright © 2019 My Math Forum. All rights reserved.