Can someone explain to me why Mathematica is ignoring the replacement rules in the following code:

eq = D[\[Psi][x, y], y] D[\[Psi][x, y], {y, 2}, {x}] - 
  D[\[Psi][x, y], x] D[\[Psi][x, y], {y, 3}] == 
  D[\[Psi][x, y], {y, 4}];
eq1 = eq /. {\[Psi][x, y] -> 
    x^\[Alpha] f[\[Eta][x, y]], \[Eta][x, y] -> y x^-\[Beta]}
eq2 = eq1 /. {\[Alpha] -> 1/2, \[Beta] -> 1/2}

Basically, I have a fourth order PDE and I want to define a similarity transformation to reduce it to a third-order ODE. Previously when I ran this, the second command was outputting something in terms of the function f and the variable \eta. However, when I now run this it is just ouputting the exact thing as eq. I don't understand the issue at all. I have tried using various clear commands, but that doesn't seem to be the issue.

As an aside, these are three lines of code in a larger notebook. However, even when I copy and past them into a different notebook the issue persists.

Can someone explain to me why Mathematica is ignoring the replacement rules in the following code:

eq = D[ψ[x, y], y] D[ψ[x, y], {y, 2}, {x}] - D[ψ[x, y], x] D[ψ[x, y], {y, 3}] == D[ψ[x, y], {y, 4}];
eq1 = eq /. {ψ[x, y] -> x^α f[η[x, y]], η[x, y] -> y x^-β}
eq2 = eq1 /. {α -> 1/2, β -> 1/2}

Basically, I have a fourth order PDE and I want to define a similarity transformation to reduce it to a third-order ODE. Previously when I ran this, the second command was outputting something in terms of the function f and the variable \eta. However, when I now run this it is just ouputting the exact thing as eq. I don't understand the issue at all. I have tried using various clear commands, but that doesn't seem to be the issue.

As an aside, these are three lines of code in a larger notebook. However, even when I copy and past them into a different notebook the issue persists.

Can someone explain to me why Mathematica is ignoring the replacement rules in the following code:

eq = D[\[Psi][x, y], y] D[\[Psi][x, y], {y, 2}, {x}] - 
  D[\[Psi][x, y], x] D[\[Psi][x, y], {y, 3}] == 
  D[\[Psi][x, y], {y, 4}];
eq1 = eq /. {\[Psi][x, y] -> 
    x^\[Alpha] f[\[Eta][x, y]], \[Eta][x, y] -> y x^-\[Beta]}
eq2 = eq1 /. {\[Alpha] -> 1/2, \[Beta] -> 1/2}

Basically, I have a fourth order PDE and I want to define a similarity transformation to reduce it to a third-order ODE. Previously when I ran this, the second command was outputting something in terms of the function f and the variable \eta. However, when I now run this it is just ouputting the exact thing as eq. I don't understand the issue at all. I have tried using various clear commands, but that doesn't seem to be the issue.

Can someone explain to me why Mathematica is ignoring the replacement rules in the following code:

eq = D[\[Psi][x, y], y] D[\[Psi][x, y], {y, 2}, {x}] - 
  D[\[Psi][x, y], x] D[\[Psi][x, y], {y, 3}] == 
  D[\[Psi][x, y], {y, 4}];
eq1 = eq /. {\[Psi][x, y] -> 
    x^\[Alpha] f[\[Eta][x, y]], \[Eta][x, y] -> y x^-\[Beta]}
eq2 = eq1 /. {\[Alpha] -> 1/2, \[Beta] -> 1/2}

Basically, I have a fourth order PDE and I want to define a similarity transformation to reduce it to a third-order ODE. Previously when I ran this, the second command was outputting something in terms of the function f and the variable \eta. However, when I now run this it is just ouputting the exact thing as eq. I don't understand the issue at all. I have tried using various clear commands, but that doesn't seem to be the issue.

As an aside, these are three lines of code in a larger notebook. However, even when I copy and past them into a different notebook the issue persists.