delvin

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These are questions asked by delvin

Hello
I have a problem in writing the Maple code of the image below, I don't know why the 3.5 answers are not available?

which one is better?

123.mw

0123.mw

Why aren't all the variables in fin 1 equation?

And the answers are different from the solutions?

 

restart

with(student)

eq1 := 12*gamma^3*rho[3]^2*(diff(w(psi), `$`(psi, 2)))+(-3*gamma*rho[2]^2+4*omega*rho[3]^2)*w(psi)+gamma*rho[3]^2*(rho[1]+2*rho[3])*w(psi)^3

12*gamma^3*rho[3]^2*(diff(diff(w(psi), psi), psi))+(-3*gamma*rho[2]^2+4*omega*rho[3]^2)*w(psi)+gamma*rho[3]^2*(rho[1]+2*rho[3])*w(psi)^3

(1)

NULL

"w(psi):=kappa[0]+sum(kappa[i]*((diff(E,psi))^(i))/((E(psi))^(i)),i=1..1)+sum(h[i]*(((diff(E,psi))^())/((E(psi))^()))^(-i),i=1..1);"

proc (psi) options operator, arrow, function_assign; kappa[0]+sum(kappa[i]*(diff(E, psi))^i/E(psi)^i, i = 1 .. 1)+sum(h[i]*((diff(E, psi))/E(psi))^(-i), i = 1 .. 1) end proc

(2)

"E(psi):=((epsilon[1]*jacobiCN(Zeta[1]*psi))+(epsilon[2]*jacobiSN(Zeta[2]*psi)))/((epsilon[3]*jacobiCN(Zeta[3]*psi))+(epsilon[4]*jacobiSN(Zeta[4]*psi))) ;"

proc (psi) options operator, arrow, function_assign; (varepsilon[1]*jacobiCN(Zeta[1]*psi)+varepsilon[2]*jacobiSN(Zeta[2]*psi))/(varepsilon[3]*jacobiCN(Zeta[3]*psi)+varepsilon[4]*jacobiSN(Zeta[4]*psi)) end proc

(3)

 

NULL

fin1 := simplify(eq1)

kappa[0]*(gamma*rho[3]^2*(rho[1]+2*rho[3])*kappa[0]^2-3*gamma*rho[2]^2+4*omega*rho[3]^2)

(4)

Sol := solve(fin1, {omega, Zeta[1], Zeta[2], Zeta[3], Zeta[4], epsilon[1], epsilon[2], epsilon[3], epsilon[4], h[1], kappa[0], kappa[1]})

{omega = omega, Zeta[1] = Zeta[1], Zeta[2] = Zeta[2], Zeta[3] = Zeta[3], Zeta[4] = Zeta[4], h[1] = h[1], kappa[0] = 0, kappa[1] = kappa[1], varepsilon[1] = varepsilon[1], varepsilon[2] = varepsilon[2], varepsilon[3] = varepsilon[3], varepsilon[4] = varepsilon[4]}, {omega = -(1/4)*gamma*(kappa[0]^2*rho[1]*rho[3]^2+2*kappa[0]^2*rho[3]^3-3*rho[2]^2)/rho[3]^2, Zeta[1] = Zeta[1], Zeta[2] = Zeta[2], Zeta[3] = Zeta[3], Zeta[4] = Zeta[4], h[1] = h[1], kappa[0] = kappa[0], kappa[1] = kappa[1], varepsilon[1] = varepsilon[1], varepsilon[2] = varepsilon[2], varepsilon[3] = varepsilon[3], varepsilon[4] = varepsilon[4]}

(5)

for i to 2 do Case[i] := allvalues(Sol[i]) end do

{omega = omega, Zeta[1] = Zeta[1], Zeta[2] = Zeta[2], Zeta[3] = Zeta[3], Zeta[4] = Zeta[4], h[1] = h[1], kappa[0] = 0, kappa[1] = kappa[1], varepsilon[1] = varepsilon[1], varepsilon[2] = varepsilon[2], varepsilon[3] = varepsilon[3], varepsilon[4] = varepsilon[4]}

 

{omega = -(1/4)*gamma*(kappa[0]^2*rho[1]*rho[3]^2+2*kappa[0]^2*rho[3]^3-3*rho[2]^2)/rho[3]^2, Zeta[1] = Zeta[1], Zeta[2] = Zeta[2], Zeta[3] = Zeta[3], Zeta[4] = Zeta[4], h[1] = h[1], kappa[0] = kappa[0], kappa[1] = kappa[1], varepsilon[1] = varepsilon[1], varepsilon[2] = varepsilon[2], varepsilon[3] = varepsilon[3], varepsilon[4] = varepsilon[4]}

(6)

NULL

NULL

Download 0123.mw

Hello

I want to write a program to get unknown coefficients of multiple polynomials. I have a problem with this program. The code sometimes doesn't work. Can anyone help me? It's very important to me.

restart

with(student)

``

EQ[0] := tanh(d)*b[1]*(b[1]+1)

tanh(d)*b[1]*(b[1]+1)

(1)

EQ[1] := -(-1+(a[1]-b[1]-1)*tanh(d)^2+(a[0]+1)*tanh(d))*b[1]

-(-1+(a[1]-b[1]-1)*tanh(d)^2+(a[0]+1)*tanh(d))*b[1]

(2)

EQ[2] := tanh(d)*((a[1]-b[1])*(a[0]+1)*tanh(d)-b[1]^2-a[1])

tanh(d)*((a[1]-b[1])*(a[0]+1)*tanh(d)-b[1]^2-a[1])

(3)

EQ[3] := (-a[1]^2+(2*b[1]-1)*a[1]-b[1]^2-b[1])*tanh(d)^2+(a[1]+b[1])*(a[0]+1)*tanh(d)-a[1]-b[1]

(-a[1]^2+(2*b[1]-1)*a[1]-b[1]^2-b[1])*tanh(d)^2+(a[1]+b[1])*(a[0]+1)*tanh(d)-a[1]-b[1]

(4)

EQ[4] := -tanh(d)*((a[1]-b[1])*(a[0]+1)*tanh(d)+a[1]^2+b[1])

-tanh(d)*((a[1]-b[1])*(a[0]+1)*tanh(d)+a[1]^2+b[1])

(5)

EQ[5] := -a[1]*(-1+(-a[1]+b[1]-1)*tanh(d)^2+(a[0]+1)*tanh(d))

-a[1]*(-1+(-a[1]+b[1]-1)*tanh(d)^2+(a[0]+1)*tanh(d))

(6)

EQ[6] := (a[1]+1)*a[1]*tanh(d)

(a[1]+1)*a[1]*tanh(d)

(7)

Eqs := {seq(EQ[i], i = 0 .. 6)}

Sol := solve(Eqs, {a[0], a[1], b[1]})NULL

{a[0] = a[0], a[1] = 0, b[1] = 0}, {a[0] = (tanh(d)^2-tanh(d)+1)/tanh(d), a[1] = -1, b[1] = -1}, {a[0] = -(tanh(d)-1)/tanh(d), a[1] = -1, b[1] = 0}, {a[0] = -(tanh(d)-1)/tanh(d), a[1] = 0, b[1] = -1}

(8)

for i from 2 to 4 do Case[i] := allvalues(Sol[i]) end do

{a[0] = -(tanh(d)-1)/tanh(d), a[1] = 0, b[1] = -1}

(9)

``

T1.mw

T2.mw

Hi,

`[Length of output exceeds limit of 1000000]`

Hello, I want to get the the homogeneous balance principle for the Differential-Difference Equation with Maple. Can anyone help?

the homogeneous balance principle:The balance is made between sentences with the highest degree of nonlinearity and the highest order of the available derivative. We consider the power of terms like u^p as pM and u(q) as M + q and put them equal (pM=M+q) and get the value of M. Now, if M = 1/n (where m is an integer), then we use the transformation U= W^n, where W is a new function.

example:

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