偏微分方程∂^2*u/∂x^2+∂^2*u/∂y^2=2(x^2+y^2)的数值解法...答:源程序:function [u,x,y] =Helmholtz(f,g,bx0,bxf,by0,byf,D,Mx,My,MinErr,MaxIter)解方程: u_xx + u_yy + g(x,y)u = f(x,y)自变量取值区域 D = [x0,xf,y0,yf] = {(x,y) |x0 <= x <= xf, y0<= y <= yf} 边界条件 u(x0,y) = bx0(y), u(xf,y) ...