@Kitonum 

tnx my friend, I followed this command(unapply). thank you. 

@Carl Love 

tnx for your answer, I know that but can you express f function in terms of zeta only? my goal is to express f function in terms of zeta only and no x in it. 

@Carl Love 

Thank you so much, but this equation has several roots, how can I get the other roots?

@tomleslie 

thank you for this point but that answer(

https://www.mapleprimes.com/questions/227667-Solve-PDE-With-RungeKutta-Method-And-Plot

) has been deleted!!!

@tomleslie 

I created the right PDE and attached again, and the boundary conditions are:

w(0,t)=0
wx(0,t)=0
wxx(L,t)=0
wxxx(L,t)=0

and the initial conditions are:

w(x,0)=0
wt(x,0)=0

PDE := 35139.16048*(diff(w(x, t), x, x, x, x))+98646.00937*(diff(w(x, t), t, t))-2771.636*(diff(w(x, t), x, x)) = 24883.20000

code.mw

@Carl Love 

nothing, there aren't points for those states, you are right, in partial derivatives we always assume that the other variables are constant, thank you, you solved my problem.

@Carl Love 

Diff(P,T)[V] is diff(P,T) when V is constant and Diff(P,V)[T] is diff(P,V) when T is constant, infact I want to obtain the diff(P,T) , diff(P,V) and diff(P,V,V) with this equation:

EQ := V^3-R*T*V^2/P-(B^2+P*B*R*T*sqrt(T)/(P-A))*V-A*B/(P*sqrt(T)) = 0;

@Carl Love 

Diff(P,T)[V] is diff(P,T) when V is constant and Diff(P,V)[T] is diff(P,V) when T is constant, infact I want to obtain the diff(P,T) , diff(P,V) and diff(P,V,V) with this equation:

EQ := V^3-R*T*V^2/P-(B^2+P*B*R*T*sqrt(T)/(P-A))*V-A*B/(P*sqrt(T)) = 0;
 

@tomleslie 

tnx for your answer, but when I used the Directsearch command the Maple has this error: "Error, `DirectSearch` does not evaluate to a module", while I installed the Directsearch package. why?

@Kitonum 

Thank you so much my friend, how can I solve it with simpson method?

with regards

@acer 

that is my project and I must solve it with code in maple or direct command in maple like ApproximateInt. 

@tomleslie 

thank you so much for your answer, how can I solve these equations with numeric and stiff=true options? 

case1:

eqs1:=diff(r[1](t), t, t)+.73732117*(diff(r[1](t), t))+1190.3671*r[1](t)-0.15306600e-2*V(t) = -0.45063508e-2*q(t)-0.13147485e-1*cos(23*t), 2.03*10^(-8)*(diff(V(t), t))+1.*10^(-6)*V(t)+0.15306600e-2*(diff(r[1](t), t)) = 0, diff(q(t), t, t)+(22.847947*(q(t)^2-1))*(diff(q(t), t))+5800.3185*q(t) = -5269.2857*(diff(r[1](t), t, t))-59.454546*cos(23*t);
ics1:=V(0) = 0, q(0) = 0, (D(q))(0) = 0, r[1](0) = 0, (D(r[1]))(0) = 0;

case2:
eqs2:=diff(r[2](t), t, t)+.41088522*(diff(r[2](t), t))+79778.138*r[2](t)-0.11624742e-1*V(t) = 0.21841836e-2*q(t)-0.50553548e-2*cos(23*t), 2.03*10^(-8)*(diff(V(t), t))+1.*10^(-6)*V(t)+0.15306600e-2*(diff(r[1](t), t))+0.11624742e-1*(diff(r[2](t), t)) = 0, diff(q(t), t, t)+(22.847947*(q(t)^2-1))*(diff(q(t), t))+5800.3185*q(t) = -5269.2857*(diff(r[1](t), t, t))+2553.9707*(diff(r[2](t), t, t))-59.454546*cos(23*t);
ics2:=V(0) = 0, q(0) = 0, (D(q))(0) = 0, r[2](0) = 0, (D(r[2]))(0) = 0;

@acer 

Hi again, I edited my equations and initial conditions, now i want to plot V(t) in terms of U0 in t=1s. tnx

eq1:=diff(r[1](t), t, t)+(.3754075596+.3045306134*U[0])*(diff(r[1](t), t))+740.0274933*r[1](t)-0.1181114312e-2*V(t) = -0.5234785518e-2*U[0]^2*q(t):
eq2:=2.030000000*10^(-8)*(diff(V(t), t))+4.065040650*10^(-7)*V(t)+0.1181114312e-2*(diff(r[1](t), t)) = 0:
eq3:=diff(q(t), t, t)-4.997988315*U[0]*(diff(q(t), t))+277.5543022*U[0]^2*q(t) = 0:
ics := V(0) = 0, q(0) = 0.1e-3, (D(q))(0) = 0, r[1](0) = 0, (D(r[1]))(0) = 0

@acer 

tnx for reply, forgive me for this wrong, the value A=1 and I just obtain r1(t), V(t) , q(t) in terms of U0, ignore the problem plot. I just need solve the ODEs in terms of U0. tnx. 

@mehdi jafari 

tnx for your answer, yes, your think is right, i have to obtain A_max where is the maximum value of the f(t) function at steady state.