EDsI[[6]]=
h[t]
4
3/2
x
-
b[x,t]h[t]
2
c
3/2
x
-
2
′
h
[t]
2
c
x
-
2
′′
Xi[2]
[t]
2
c
+
2Xi[2][t]
(0,1)
b
[x,t]
2
c
x
+
2h[t]
(1,0)
b
[x,t]
2
c
x
+
2
′
Xi[2]
[t]
(1,0)
b
[x,t]
2
c
+2
(1,0)
f
[x,t]
In[]:=
(*Fromvseparatethecoefficientsofrpolynomials.*)​​(*
0
x
:
2
′
Xi[2]
[t]
(1,0)
b
[x,t]
2
c
-
2
′′
Xi[2]
[t]
2
c
+2
(1,0)
f
[x,t]==0*)​​(*
-3/2
x
:
h[t]
4
-
b[x,t]h[t]
2
c
==0,​​(*
-1/2
x
:-
2
′
h
[t]
2
c
+
2h[t]
(1,0)
b
[x,t]
2
c
0*)​​(*
-1
x
:
2Xi[2][t]
(0,1)
b
[x,t]
2
c
0*)​​weusetheseequationsforfurthercomputation*)​​(*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}*)​​DSolve-
2
′
h
[t]
2
c
+
2h[t]
(1,0)
b
[x,t]
2
c
0/.b[x,t]->
c
4
,
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/.b[x,t]->
c
4
,{h[t],Xi[2][t]},{t},{f[x,t]},{x,t}
DSolve
:The function h[t] appears with no arguments.
Out[]=
DSolve-
2
′
h
[t]
2
c
+
2h[t]
(1,0)
b
[x,t]
2
c
0,
h[t]
4
-
h[t]
4c
0,
2Xi[2][t]
(0,1)
b
[x,t]
2
c
0,{h[t],Xi[2][t]},{t},{f[x,t]},{x,t}
​​