Substituting inside a differential equation

Question

x y''[x]+2y'[x]+l^2 x y[x]==0/.y[x]->Cos[x]

I think above code shows what I want to do quite clearly. How do I make it happen. Only y[x] is replaced.

Answer 1

This question has been asked for times, Evaluation of Derivative in a Module and Replace rule with function? Derivatives don't evaluate.

Here are several ways to solve it:

If you do not mind pollute Global namespace, assign the function first.

y[x_] = Cos[x]
x y''[x] + 2 y'[x] + l^2 x y[x] == 0

If you mind, assign it in a Block

Block[{y}, y[x_] = Cos[x];   
x y''[x] + 2 y'[x] + l^2 x y[x] == 0]

or, replace it by a pure function:

x y''[x] + 2 y'[x] + l^2 x y[x] == 0 /. y -> Function[x, Cos[x]]

If you are trying to do something more complex, then these methods will fail:

In[1]:= F[func_] := Block[{y}, y[x_] = func;
  x y''[x] + 2 y'[x] + l^2 x y[x] == 0]

In[2]:= F[Cos[x]]

Out[2]= l^2 x Cos[x] == 0

and

In[1]:= F[func_] := x y''[x] + 2 y'[x] + l^2 x y[x] == 0 /. y -> Function[x, func]

In[2]:= F[Cos[x]]

Out[2]= l^2 x Cos[x] == 0

The methods in the link will work:

In[7]:= F[func_] := 
 Block[{y, e}, e = x y''[x] + 2 y'[x] + l^2 x y[x] == 0;
    e /. y -> Function[x, #]] &[func]

In[8]:= F[Cos[x]]

Out[8]= -x Cos[x] + l^2 x Cos[x] - 2 Sin[x] == 0

or

In[11]:= F[func_] := 
 x y''[x] + 2 y'[x] + l^2 x y[x] == 0 /. y -> Function[x, #] &[func]

In[12]:= F[Cos[x]]

Out[12]= -x Cos[x] + l^2 x Cos[x] - 2 Sin[x] == 0

I do not quite understand why the last two methods work, I hope someone can explain it in this answer.