# Get the solution of the following differential equation?

Posted 5 months ago
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 I am dealing with current problems and solving differential equations.phi1 [x] was solved using DSolve.However, I used DSolve to solve phi2 [x] using 4 boundary conditions, but instead of outputting results, It just show my input.I write the relevant code below(OutPut for phi2 [x] is omitted) and attach the Mathemtica file.What's wrong? ClearAll[phi1, phi2] eqn1 := D[phi1[x], {x, 2}] == -q nd / esi bc1 := phi1[tch + tox esi/eox] == vg - vfb bc2 := phi1[-tox esi/eox] == vg - vfb DSolve[{eqn1, bc1, bc2}, phi1[x], x] {{phi1[x] -> ( eox esi nd q tch tox + esi^2 nd q tox^2 - 2 eox^2 esi vfb + 2 eox^2 esi vg + eox^2 nd q tch x - eox^2 nd q x^2)/( 2 eox^2 esi)}} eqn2 := D[phi2[x, y], {x, 2}] + D[phi2[x, y], {y, 2}] == 0 bc3 := phi2[x, 0] == vs - phi1[x] bc4 := phi2[x, l] == vs + vd - phi1[x] bc5 := phi2[tch + tox esi/eox, y] == 0 bc6 := phi2[-tox esi/eox, y] == 0 DSolve[{eqn2, bc3, bc4, bc5, bc6}, phi2[x, y], {x, y}]  Attachments:
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Posted 5 months ago
 I'll give you a very similar case In[1]:= s = DSolveValue[{v''[x] == -2*a, v[0] == v0, v[x0] == v0}, v, x] Out[1]= Function[{x}, v0 - a x^2 + a x x0] In[2]:= bc = {u[x, 0] == u0 + s[x], u[x, y0] == u1 + s[x], u[0, y] == 0, u[x0, y] == 0}; leqn = Laplacian[u[x, y], {x, y}] == 0; In[4]:= sol = FullSimplify[ DSolveValue[{leqn, bc}, u[x, y], {x, y}, Assumptions -> x0 > 0]] Out[4]= Inactive[Sum][( 4 Csch[(\[Pi] y0 K[1])/x0] Sin[1/2 \[Pi] K[1]]^2 Sin[(\[Pi] x K[1])/ x0] ((2 a x0^2 + \[Pi]^2 (u1 + v0) K[1]^2) Sinh[(\[Pi] y K[1])/ x0] + (2 a x0^2 + \[Pi]^2 (u0 + v0) K[1]^2) Sinh[(\[Pi] (-y + y0) K[1])/x0]))/(\[Pi]^3 K[1]^3), {K[1], 1, \[Infinity]}] 
Posted 5 months ago
 Thank you. However, Due to my lack of knowledge, I do not know what's wrong.
Posted 5 months ago
 The function phi1[x] is not defined in the boundary conditions: bc3 := phi2[x, 0] == vs - phi1[x] bc4 := phi2[x, l] == vs + vd - phi1[x] 
Posted 5 months ago
 Thanks. I'll try to solve it.