How to solve an ODE with power function coefficient?

Questions
,
1

x^(2 (1 + [Alpha]) ) D[u[x, t],{ x，2}] - t u[x, t] == 0

2

How about DSolve:

DSolve[x^(2 (1 + \[Alpha])) D[u[x, t], x] - t u[x, t] == 0,
 u[x, t], x]
3

The former input was wrong, I modified it and the equation is still solvable.

4

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]
5

Oh, that’s too cute! +1 :slight_smile: :slight_smile:

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.)

6

Brilliant. That’s the real trouble confusing me.