The next thing you need to do, aside from supplying numerical values to your variables, is to give your rule a name. Let's just call it ruel for now:
In[5]:= ruel =
DSolve[{Derivative[2][Subscript[u, 0, 0]][y] +
Rey*S*Derivative[1][Subscript[u, 0, 0]][y] == -Rey*
Gr*((E^(Pr Rey S - Pr Rey S y) (-1 + E^(Pr Rey S y)))/(-1 +
E^(Pr Rey S))),
Subscript[u, 0, 0][0] == h*Derivative[1][Subscript[u, 0, 0]][0],
Subscript[u, 0, 0][1] == 1}, Subscript[u, 0, 0][y], y] //
FullSimplify
Out[5]= {{Subscript[u, 0, 0][
y] -> (E^(-Rey S (Pr (-1 + y) + y)) (-E^(Rey S y) Gr +
E^(Rey S (1 + y)) Gr (1 + h Rey S) +
E^(Rey S (1 + Pr (-1 + y))) (Gr + (-1 + Pr) Pr Rey S^2) -
E^(Rey S (1 + Pr (-1 + y) + y)) (1 +
h Rey S) (Gr + (-1 + Pr) Pr Rey S^2) -
E^(Rey S (1 + Pr y)) (Gr +
Gr Pr (-1 + Pr + h Pr) Rey S + (-1 + Pr) Pr Rey S^2) +
E^(Rey S (1 + y + Pr y)) (-1 + Pr) Pr Rey S (1 +
h Rey S) (Gr + S - Gr y) +
E^((1 + Pr) Rey S y)
Gr (1 + Pr Rey S (h Pr + (-1 + Pr) y))))/((-1 + E^(
Pr Rey S)) (-1 + Pr) Pr Rey S^2 (-1 +
E^(Rey S) (1 + h Rey S)))}}
What happens when we just print ruel then? Let's find out:
In[6]:= ruel
Out[6]= {{Subscript[u, 0, 0][
y] -> (E^(-Rey S (Pr (-1 + y) + y)) (-E^(Rey S y) Gr +
E^(Rey S (1 + y)) Gr (1 + h Rey S) +
E^(Rey S (1 + Pr (-1 + y))) (Gr + (-1 + Pr) Pr Rey S^2) -
E^(Rey S (1 + Pr (-1 + y) + y)) (1 +
h Rey S) (Gr + (-1 + Pr) Pr Rey S^2) -
E^(Rey S (1 + Pr y)) (Gr +
Gr Pr (-1 + Pr + h Pr) Rey S + (-1 + Pr) Pr Rey S^2) +
E^(Rey S (1 + y + Pr y)) (-1 + Pr) Pr Rey S (1 +
h Rey S) (Gr + S - Gr y) +
E^((1 + Pr) Rey S y)
Gr (1 + Pr Rey S (h Pr + (-1 + Pr) y))))/((-1 + E^(
Pr Rey S)) (-1 + Pr) Pr Rey S^2 (-1 +
E^(Rey S) (1 + h Rey S)))}}
Now we have given the rule a name, we can (this would my preference) strip away the outer brackets by taking the innermost part of the rule, like so:
In[7]:= ruel[[1, 1]]
Out[7]= Subscript[u, 0, 0][
y] -> (E^(-Rey S (Pr (-1 + y) + y)) (-E^(Rey S y) Gr +
E^(Rey S (1 + y)) Gr (1 + h Rey S) +
E^(Rey S (1 + Pr (-1 + y))) (Gr + (-1 + Pr) Pr Rey S^2) -
E^(Rey S (1 + Pr (-1 + y) + y)) (1 +
h Rey S) (Gr + (-1 + Pr) Pr Rey S^2) -
E^(Rey S (1 + Pr y)) (Gr +
Gr Pr (-1 + Pr + h Pr) Rey S + (-1 + Pr) Pr Rey S^2) +
E^(Rey S (1 + y + Pr y)) (-1 + Pr) Pr Rey S (1 + h Rey S) (Gr +
S - Gr y) +
E^((1 + Pr) Rey S y)
Gr (1 + Pr Rey S (h Pr + (-1 + Pr) y))))/((-1 + E^(
Pr Rey S)) (-1 + Pr) Pr Rey S^2 (-1 + E^(Rey S) (1 + h Rey S)))
That would be my preference in making use of a rule.
Let's make use of our rule in replacing the original variable u 00 [ y ] by using the replace part symbol
/.:
In[8]:= Subscript[u, 0, 0][y] /. ruel[[1, 1]]
Out[8]= (E^(-Rey S (Pr (-1 + y) + y)) (-E^(Rey S y) Gr +
E^(Rey S (1 + y)) Gr (1 + h Rey S) +
E^(Rey S (1 + Pr (-1 + y))) (Gr + (-1 + Pr) Pr Rey S^2) -
E^(Rey S (1 + Pr (-1 + y) + y)) (1 +
h Rey S) (Gr + (-1 + Pr) Pr Rey S^2) -
E^(Rey S (1 + Pr y)) (Gr +
Gr Pr (-1 + Pr + h Pr) Rey S + (-1 + Pr) Pr Rey S^2) +
E^(Rey S (1 + y + Pr y)) (-1 + Pr) Pr Rey S (1 + h Rey S) (Gr +
S - Gr y) +
E^((1 + Pr) Rey S y)
Gr (1 + Pr Rey S (h Pr + (-1 + Pr) y))))/((-1 + E^(
Pr Rey S)) (-1 + Pr) Pr Rey S^2 (-1 + E^(Rey S) (1 + h Rey S)))
We still haven't set u 00 [ y ] equal to anything, but once we do then it will only be necessary to supply some numbers to your variables and hopefully your problem will be ready to Plot.
You could use the rule and replace part ( this symbol, the slash and dot ? /. is known as replace part ) in plotting, or you could set u 0,0 [ y ] equal to the material indicated in the rule.
Here we set u 0,0 [ y ] equal to the material indicated in the rule:
In[9]:= Subscript[u, 0, 0][y] = Subscript[u, 0, 0][y] /. ruel[[1, 1]]
Out[9]= (E^(-Rey S (Pr (-1 + y) + y)) (-E^(Rey S y) Gr +
E^(Rey S (1 + y)) Gr (1 + h Rey S) +
E^(Rey S (1 + Pr (-1 + y))) (Gr + (-1 + Pr) Pr Rey S^2) -
E^(Rey S (1 + Pr (-1 + y) + y)) (1 +
h Rey S) (Gr + (-1 + Pr) Pr Rey S^2) -
E^(Rey S (1 + Pr y)) (Gr +
Gr Pr (-1 + Pr + h Pr) Rey S + (-1 + Pr) Pr Rey S^2) +
E^(Rey S (1 + y + Pr y)) (-1 + Pr) Pr Rey S (1 + h Rey S) (Gr +
S - Gr y) +
E^((1 + Pr) Rey S y)
Gr (1 + Pr Rey S (h Pr + (-1 + Pr) y))))/((-1 + E^(
Pr Rey S)) (-1 + Pr) Pr Rey S^2 (-1 + E^(Rey S) (1 + h Rey S)))
What happens when we just write down u 0,0 [ y ]? Let's take a look:
In[10]:= Subscript[u, 0, 0][y]
Out[10]= (E^(-Rey S (Pr (-1 + y) + y)) (-E^(Rey S y) Gr +
E^(Rey S (1 + y)) Gr (1 + h Rey S) +
E^(Rey S (1 + Pr (-1 + y))) (Gr + (-1 + Pr) Pr Rey S^2) -
E^(Rey S (1 + Pr (-1 + y) + y)) (1 +
h Rey S) (Gr + (-1 + Pr) Pr Rey S^2) -
E^(Rey S (1 + Pr y)) (Gr +
Gr Pr (-1 + Pr + h Pr) Rey S + (-1 + Pr) Pr Rey S^2) +
E^(Rey S (1 + y + Pr y)) (-1 + Pr) Pr Rey S (1 + h Rey S) (Gr +
S - Gr y) +
E^((1 + Pr) Rey S y)
Gr (1 + Pr Rey S (h Pr + (-1 + Pr) y))))/((-1 + E^(
Pr Rey S)) (-1 + Pr) Pr Rey S^2 (-1 + E^(Rey S) (1 + h Rey S)))
You could
probably Plot that if only those numbers meant something.
I say
probably, because without knowing if the formula would give purely real results or complex ones I can't say for sure. If you supply numbers which end up giving complex results, it still won't Plot and you will have to decide which part you want to look at, the real part Re[ ] or the imaginary part Im[ ].