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  eqn2 = -dp/L + \[Mu]*D[v[x, y], x, x] - 
        Subscript[\[Sigma], e]*Subscript[B, x]^2*v[x, y] + 
        Subscript[\[Sigma], e]*Subscript[E, z]*Subscript[B, x] + \[Zeta] Cosh[x \[Kappa]] Sech[\[Kappa]]*
         Subscript[E, y] == 0;
    bc2 = { v[a, y] == 0, Derivative[0, 1][v][0, y] == 0};
    sol2 = DSolveValue[{eqn2, bc2}, v[x, y], {x, y}];
    Assuming[{a > 0, \[Kappa] > 0}, Simplify[sol2]]

I am not able to get the solution

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  • 4
    $\begingroup$ There are a number of problems here. There is only one DE, but two independent variables. There are not enough boundary conditions for the DE. Using the independent variable $x$ as a subscript does not do what you think it does. Best practice is to not use subscripts in variable names. If you edit your question and add more information and an explanation of what you are trying to do, you will be more likely to get help. $\endgroup$
    – LouisB
    Commented Jan 13 at 18:31

1 Answer 1

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Hint.

Try to solve first

eqn2 = -dp/L + \[Mu]*D[v[x, y], x, x] - 
    Subscript[\[Sigma], e]*Subscript[B, x0]^2*v[x, y] + 
    Subscript[\[Sigma], e]*Subscript[E, z]*
    Subscript[B, x0] + \[Zeta] Cosh[x \[Kappa]] Sech[\[Kappa]]*
    Subscript[E, y0] == 0;
bc2 = {v[a, y] == 0 (***,Derivative[0,1][v][0,y]==0 ***)};
sol2 = DSolveValue[{eqn2, bc2}, v[x, y], {x, y}] // FullSimplify

Note that now we are using Subscript[B, x0] and Subscript[E, y0] instead of Subscript[B, x] and Subscript[E, y] and also considering only the first boundary condition. The second boundary condition can be handled after that imposing

D[(sol2/.{x -> 0}), y] == 0
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