# How to solve an ODE with power function coefficient?

Posted 5 days ago
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 x^(2 (1 + \[Alpha]) ) D[u[x, t],{ x，2}] - t u[x, t] == 0 
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Posted 5 days ago
 How about DSolve: DSolve[x^(2 (1 + \[Alpha])) D[u[x, t], x] - t u[x, t] == 0, u[x, t], x] 
Posted 5 days ago
 The former input was wrong, I modified it and the equation is still solvable.
Posted 5 days ago
 Here is a workaround: eq = x^(2 (1 + \[Alpha])) D[u[x], {x, 2}] - t u[x]; sol = Simplify[ Block[{\[Alpha] = E}, DSolve[eq == 0, u, x] // First] /. E -> \[Alpha]] FullSimplify[eq /. sol, t > 0 && x > 0] 
Posted 5 days ago
 Oh, that's too cute! +1 :) :)In general, I'd suggest using another transcendental constant, one that is less likely to show up in the solution of a differential equation so that the back substitution E -> α doesn't clobber an exponential function. (Examples: EulerGamma, Catalan, Zeta[7],..., but not Pi obviously though it, too, will work here.)
Posted 4 days ago
 Brilliant. That's the real trouble confusing me.