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 Author Comment/Response Bill Simpson 01/22/13 10:01pm Find the general solution In[1]:= pde=(1/z)D[T[n,z],z]+D[T[n,z],{z,2}]==0; sol=T[n,z]/.DSolve[pde,T[n,z],{n,z}][[1]] Out[2]= Log[z]C[1][n]+C[2][n] Take the general solution, find second derivative, then substitute for z and finally equate to zero In[3]:= Simplify[(D[sol,{z,2}]/.z->5)==0] Out[3]= C[1][n]==0 Take that information and substitute back into the general solution to find the particular solution. In[4]:= sol/.C[1][n]->0 Out[4]= C[2][n] Substitute the particular solution back into your original PDE to verify it. In[5]:= (1/z)D[C[2][n],z]+D[C[2][n],{z,2}]==0 Out[5]= True Please check all this carefully URL: ,

 Subject (listing for 'Boundary conditions for PDE') Author Date Posted Boundary conditions for PDE Teresa 01/21/13 9:53pm Re: Boundary conditions for PDE Bill Simpson 01/22/13 10:01pm
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