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The problem is

Find two linearly independent solutions of $2x^2y'' -xy' + (-4 x + 1)y = 0, \, x>0$ of the form $y_1 = x^{r_1}(1 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots)$ and $y_2 = x^{r_2}(1 + b_1 x + b_2 x^2 + b_3 x^3 + \cdots)$.

What I have done is below:

2 x^2*D[y, {x, 2}] - x D[y, x] + (-4 x + 1) y == 0 /. y -> x^r*(Sum[C[n]*x^n, {n, 0, 5}])

The result mathematica gave is

(1 - 4 x) x^r (C[0] + x C[1] + x^2 C[2] + x^3 C[3] + x^4 C[4] + x^5 C[5]) == 0

from which I conclude that D[y,x]is excuted before the replacement. How can I make D[y,x] excute after the replacement?

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  • $\begingroup$ With[{y=replacement},D[y,x]] $\endgroup$ – Coolwater Apr 27 '14 at 9:03
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One method I like to use for this is the Hold/Release, like this (notice the use of D for derivative in the Hold stage.

  ClearAll[x, y, n];
  eq = 2 x^2  Hold[D[y[x], {x, 2}]] - x Hold[D[y[x], x]] + (-4 x + 1) y[x] == 0;
  Release[eq /. y[x] -> x^r*(Sum[C[n]*x^n, {n, 0, 5}])]

Mathematica graphics

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  solns = y[x] /. First@DSolve[2 x^2*D[y[x], {x, 2}] - x D[y[x], x] + (-4 x + 1) y[x]== 0, y[x], x]
  y1[x_] = solns /. {C[1] -> 1, C[2] -> 0}
  y2[x_] = solns /. {C[1] -> 0, C[2] -> 1}
(*  E^(2 Sqrt[2] Sqrt[x]) Sqrt[x]  *)
(* -((E^(-2 Sqrt[2] Sqrt[x]) Sqrt[x])/(2 Sqrt[2]))  *)
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