If:
b[x,t]->
c
4
then:
(1,0)
b
[x,t]
=0 and
(0,1)
b
[x,t]
=0
In[]:=
-
2
′
h
[t]
2
c
+
2h[t]
(1,0)
b
[x,t]
2
c
0/.
(1,0)
b
[x,t]->0,
h[t]
4
-
b[x,t]h[t]
2
c
==0/.b[x,t]->
c
4
,
2Xi[2][t]
(0,1)
b
[x,t]
2
c
0/.
(0,1)
b
[x,t]->0
Out[]=
-
2
′
h
[t]
2
c
0,
h[t]
4
-
h[t]
4c
0,True
In[]:=
Solve
h[t]
4
-
h[t]
4c
0,h[t]
Out[]=
{{h[t]0}}
In[]:=
DSolve-
2
′
h
[t]
2
c
0,h[t],t
Out[]=
{{h[t]

1
}}
(*----------------------------------------------------------------------------------------------------------------------------*)
DSolve
2
′
Xi[2]
[t]
(1,0)
b
[x,t]
2
c
-
2
′′
Xi[2]
[t]
2
c
+2
(1,0)
f
[x,t]==0,-
2
′
h
[t]
2
c
+
2h[t]
(1,0)
b
[x,t]
2
c
0,
h[t]
4
-
b[x,t]h[t]
2
c
==0,{Xi[2][t],h[t],f[x,t],b[x,t]},{x,t}(*Syntaxiswrong!!Imeanthis:"{Xi[2][t],h[t],f[x,t],b[x,t]},{x,t}"*)(*Inthisexampleyouhave3equationsbut4unknowns?*)