Sure, I can provide an example for solving this equation in Mathematica. First, we need to define the variables and constants:
A = 2; (*constant*)
B[x_] := x^2; (*function of x*)
U[x_, y_] := u[x, y]; (*defining U as a function of x and y*)
Next, we can use the built-in function DSolve to solve the equation:
DSolve[A*(D[U[x, y], {x, 2}] + D[U[x, y], {y, 2}]) - B[x]*D[U[x, y], x] == 0, U[x, y], {x, y}]
This will give us the general solution to the equation:
{{U[x, y] -> C[1] + C[2]*y + (A*Integrate[B[x], x] + C[3])*x + (A*Integrate[B[x], x, x] + C[4])*y^2}}
Next, we can use the boundary conditions to solve for the constants C[1], C[2], C[3], and C[4]. In this case, we have two boundary conditions at x=0 and three unknown constants, so we will need to use a numerical method to solve for the constants. For example, we can use the finite difference method:
(*creating a grid for x and y*)
xgrid = Table[i, {i, -1, 1, 0.1}];
ygrid = Table[j, {j, -1, 1, 0.1}];
(*creating a matrix for the finite difference method*)
matrix = Table[0, {i, Length[xgrid]}, {j, Length[ygrid]}];
(*filling in the matrix with the equation*)
Do[
If[i == 1, matrix[[i, j]] = D[U[xgrid[], ygrid[[j]]], y]; (*boundary condition at x=0*)
matrix[[i, j]] = 0; (*boundary condition at y=0*)
,
matrix[[i, j]] = A*(D[U[xgrid[], ygrid[[j]]], {x, 2}] + D[U[xgrid[], ygrid[[j]]], {y, 2}]) - B[xgrid[]]*D[U[xgrid[[i