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MaplePrimes Questions

Why the subs command not apply ?...

Asked by: salim-barzani 1555
radical substitution subs
April 29 2025
3  9

i did every thing coreectly but nothing happen not apply where is my mistake?

> 

``
> 

restart
> 

with(PDEtools)
> 

with(LinearAlgebra)
> 

with(Physics)
> 

with(SolveTools)
> 

undeclare(prime)

`There is no more prime differentiation variable; all derivatives will be displayed as indexed functions`

 (1)
> 

NULL
> 

S := (diff(G(xi), xi))^2-r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2) = 0

(diff(G(xi), xi))^2-r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2) = 0

 (2)
> 

SS := diff(G(xi), xi) = sqrt(r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2))

diff(G(xi), xi) = (r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2))^(1/2)

 (3)
> 

Se := sqrt(r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2)) = diff(G(xi), xi)

(r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2))^(1/2) = diff(G(xi), xi)

 (4)
> 

dub := diff(SS, xi)

diff(diff(G(xi), xi), xi) = (1/2)*(2*r^2*G(xi)*(a+b*G(xi)+l*G(xi)^2)*(diff(G(xi), xi))+r^2*G(xi)^2*(b*(diff(G(xi), xi))+2*l*G(xi)*(diff(G(xi), xi))))/(r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2))^(1/2)

 (5)
> 

Dubl2 := simplify(diff(diff(G(xi), xi), xi) = (1/2)*(2*r^2*G(xi)*(a+b*G(xi)+l*G(xi)^2)*(diff(G(xi), xi))+r^2*G(xi)^2*(b*(diff(G(xi), xi))+2*l*G(xi)*(diff(G(xi), xi))))/(r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2))^(1/2))

diff(diff(G(xi), xi), xi) = (1/2)*r^2*G(xi)*(diff(G(xi), xi))*(4*l*G(xi)^2+3*b*G(xi)+2*a)/(r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2))^(1/2)

 (6)
> 

subs(SA, Dubl2)

diff((r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2))^(1/2), xi) = (1/2)*r^2*G(xi)*(4*l*G(xi)^2+3*b*G(xi)+2*a)

 (7)
> 

subs(Se, Dubl2)

diff(diff(G(xi), xi), xi) = (1/2)*r^2*G(xi)*(diff(G(xi), xi))*(4*l*G(xi)^2+3*b*G(xi)+2*a)/(r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2))^(1/2)

 (8)
> 

subs(lhs(Se) = rhs(Se), Dubl2)

diff(diff(G(xi), xi), xi) = (1/2)*r^2*G(xi)*(diff(G(xi), xi))*(4*l*G(xi)^2+3*b*G(xi)+2*a)/(r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2))^(1/2)

 (9)
 

NULL

Download subs.mw

Computing the left and right eigenvectors...

Asked by: Elisha 50
eigenvector
April 29 2025
2  1

Someone please help me with the computation of the right and left eigenvectors. my system of equation is attached below

> 

with(VectorCalculus):
 
> 

interface(imaginaryunit = I)

I

 (2)
> 

I

I

 (3)
> 

sqrt(-4)

2*I

 (4)
> 

NULL
> 

``

Limit(N(t) = N__0*exp(-mu*t)+exp(mu*t)*K/mu, t = infinity)

  

limit(N(t), t = infinity) = limit(N__0*exp(-mu*t)+exp(mu*t)*K/mu, t = infinity)

 (5)
> 

 

> 

NULL
> 

#to calculate the  disease free equilibrium,
> 

NULL
> 

E1 := -S*µ__C+`Λ__p`

-S*µ__C+Lambda__p

 (6)
> 

NULL

``

 (7)
> 

E3 := -S__A*µ__A+`Λ__A`

-S__A*µ__A+Lambda__A

 (8)
> 

NULL

``

 (9)
> 

NULL

``

 (10)
> 

NULL
> 

solve({E1 = 0, E3 = 0}, {S, S__A})

{S = Lambda__p/µ__C, S__A = Lambda__A/µ__A}

 (11)
> 

NULL
> 

NULL#to calculate the Endemic Equilibrium state,

Typesetting:-mparsed()

 (12)
> 

restart
> 

with(VectorCalculus):
 
> 

interface(imaginaryunit = I)

I

 (14)
> 

I

I

 (15)
> 

sqrt(-4)

2*I

 (16)
> 

``
> 

E1 := `Λ__p`-(`ϕ`*`θ__B`*I__A/N__p+µ__C)*S+`ω__B`*I__B

Lambda__p-(varphi*theta__B*I__A/N__p+µ__C)*S+omega__B*I__B

 (17)
> 

E2 := `ϕ`*`θ__B`*I__A*S/N__p-`ω__B`*I__B-(`σ__B`+µ__C)*I__B

varphi*theta__B*I__A*S/N__p-omega__B*I__B-(sigma__B+µ__C)*I__B

 (18)

``

 (19)
> 

E3 := `Λ__A`-(µ__A+`ϕ`*`α__B`*I__B/N__p)*S__A+`δ__A`*I__A

Lambda__A-(µ__A+varphi*alpha__B*I__B/N__p)*S__A+delta__A*I__A

 (20)
> 

E4 := `ϕ`*`α__B`*I__B*S__A/N__p-(µ__A+`δ__A`)*I__A

varphi*alpha__B*I__B*S__A/N__p-(µ__A+delta__A)*I__A

 (21)
> 

NULL

``

 (22)
> 

NULL

``

 (23)
> 

solve({E1 = 0, E2 = 0, E3 = 0, E4 = 0}, {I__A, I__B, S, S__A})

{I__A = 0, I__B = 0, S = Lambda__p/µ__C, S__A = Lambda__A/µ__A}, {I__A = -(N__p^2*µ__A^2*µ__C^2+N__p^2*µ__A^2*µ__C*omega__B+N__p^2*µ__A^2*µ__C*sigma__B+N__p^2*µ__A*µ__C^2*delta__A+N__p^2*µ__A*µ__C*delta__A*omega__B+N__p^2*µ__A*µ__C*delta__A*sigma__B-varphi^2*Lambda__A*Lambda__p*alpha__B*theta__B)/(varphi*µ__A*theta__B*(N__p*µ__A*µ__C+N__p*µ__A*sigma__B+N__p*µ__C*delta__A+N__p*delta__A*sigma__B+varphi*Lambda__p*alpha__B)), I__B = -(N__p^2*µ__A^2*µ__C^2+N__p^2*µ__A^2*µ__C*omega__B+N__p^2*µ__A^2*µ__C*sigma__B+N__p^2*µ__A*µ__C^2*delta__A+N__p^2*µ__A*µ__C*delta__A*omega__B+N__p^2*µ__A*µ__C*delta__A*sigma__B-varphi^2*Lambda__A*Lambda__p*alpha__B*theta__B)/(alpha__B*(N__p*µ__A*µ__C^2+N__p*µ__A*µ__C*omega__B+N__p*µ__A*µ__C*sigma__B+varphi*µ__C*Lambda__A*theta__B+varphi*Lambda__A*sigma__B*theta__B)*varphi), S = (N__p*µ__A*µ__C+N__p*µ__A*sigma__B+N__p*µ__C*delta__A+N__p*delta__A*sigma__B+varphi*Lambda__p*alpha__B)*µ__A*N__p*(µ__C+omega__B+sigma__B)/(alpha__B*varphi*(N__p*µ__A*µ__C^2+N__p*µ__A*µ__C*omega__B+N__p*µ__A*µ__C*sigma__B+varphi*µ__C*Lambda__A*theta__B+varphi*Lambda__A*sigma__B*theta__B)), S__A = N__p*(N__p*µ__A^2*µ__C^2+N__p*µ__A^2*µ__C*omega__B+N__p*µ__A^2*µ__C*sigma__B+N__p*µ__A*µ__C^2*delta__A+N__p*µ__A*µ__C*delta__A*omega__B+N__p*µ__A*µ__C*delta__A*sigma__B+varphi*µ__A*µ__C*Lambda__A*theta__B+varphi*µ__A*Lambda__A*sigma__B*theta__B+varphi*µ__C*Lambda__A*delta__A*theta__B+varphi*Lambda__A*delta__A*sigma__B*theta__B)/(varphi*µ__A*theta__B*(N__p*µ__A*µ__C+N__p*µ__A*sigma__B+N__p*µ__C*delta__A+N__p*delta__A*sigma__B+varphi*Lambda__p*alpha__B))}

 (24)
> 

``

J := Jacobian([E1, E2, E3, E4], [S, I__B, S__A, I__A])

Matrix(%id = 18446746854857131062)

 (25)

NULL

> 

restart
> 

J := Matrix(4, 4, {(1, 1) = -`ϕ`*`θ__B`*I__A/N__p-µ__C, (1, 2) = `ω__B`, (1, 3) = 0, (1, 4) = -`ϕ`*`θ__B`*S/N__p, (2, 1) = `ϕ`*`θ__B`*I__A/N__p, (2, 2) = -`ω__B`-`σ__B`-µ__C, (2, 3) = 0, (2, 4) = `ϕ`*`θ__B`*S/N__p, (3, 1) = 0, (3, 2) = -`ϕ`*`α__B`*S__A/N__p, (3, 3) = -µ__A-`ϕ`*`α__B`*I__B/N__p, (3, 4) = `δ__A`, (4, 1) = 0, (4, 2) = `ϕ`*`α__B`*S__A/N__p, (4, 3) = `ϕ`*`α__B`*I__B/N__p, (4, 4) = -µ__A-`δ__A`})

Matrix(%id = 18446746579340105118)

 (26)

S := `Λ__p`/µ__C

Lambda__p/µ__C

 (27)

S__A := `Λ__A`/µ__A

Lambda__A/µ__A

 (28)

I__B := 0

0

 (29)

I__A := 0

0

 (30)

NULL

0

 (31)

J := Matrix(4, 4, {(1, 1) = -`ϕ`*`θ__B`*I__A/N__p-µ__C, (1, 2) = `ω__B`, (1, 3) = 0, (1, 4) = -`ϕ`*`θ__B`*S/N__p, (2, 1) = `ϕ`*`θ__B`*I__A/N__p, (2, 2) = -`ω__B`-`σ__B`-µ__C, (2, 3) = 0, (2, 4) = `ϕ`*`θ__B`*S/N__p, (3, 1) = 0, (3, 2) = -`ϕ`*`α__B`*S__A/N__p, (3, 3) = -µ__A-`ϕ`*`α__B`*I__B/N__p, (3, 4) = `δ__A`, (4, 1) = 0, (4, 2) = `ϕ`*`α__B`*S__A/N__p, (4, 3) = `ϕ`*`α__B`*I__B/N__p, (4, 4) = -µ__A-`δ__A`})

Matrix(%id = 18446746579340107518)

 (32)
> 

J := Matrix(4, 4, {(1, 1) = -µ__C, (1, 2) = `ω__B`, (1, 3) = 0, (1, 4) = -`β__1`, (2, 1) = 0, (2, 2) = -`ω__B`-`σ__B`-µ__C, (2, 3) = 0, (2, 4) = -`β__1`, (3, 1) = 0, (3, 2) = -`β__2`, (3, 3) = -µ__A, (3, 4) = `δ__A`, (4, 1) = 0, (4, 2) = `β__2`, (4, 3) = 0, (4, 4) = -µ__A-`δ__A`})

Matrix(%id = 18446746579417403630)

 (33)
> 

"simplify( ? )"

Matrix(%id = 18446746579305905318)

 (34)
> 

"LinearAlgebra:-Eigenvalues( ? )"

Vector[column](%id = 18446746579445964182)

 (35)
> 

"LinearAlgebra:-CharacteristicPolynomial( ?, lambda )"

lambda^4+(2*µ__A+delta__A+omega__B+sigma__B+2*µ__C)*lambda^3+(beta__1*beta__2+µ__A^2+4*µ__A*µ__C+µ__A*delta__A+2*µ__A*omega__B+2*µ__A*sigma__B+µ__C^2+2*µ__C*delta__A+µ__C*omega__B+µ__C*sigma__B+delta__A*omega__B+delta__A*sigma__B)*lambda^2+(beta__1*beta__2*µ__A+beta__1*beta__2*µ__C+2*µ__A^2*µ__C+µ__A^2*omega__B+µ__A^2*sigma__B+2*µ__A*µ__C^2+2*µ__A*µ__C*delta__A+2*µ__A*µ__C*omega__B+2*µ__A*µ__C*sigma__B+µ__A*delta__A*omega__B+µ__A*delta__A*sigma__B+µ__C^2*delta__A+µ__C*delta__A*omega__B+µ__C*delta__A*sigma__B)*lambda+beta__1*beta__2*µ__A*µ__C+µ__A^2*µ__C^2+µ__A^2*µ__C*omega__B+µ__A^2*µ__C*sigma__B+µ__A*µ__C^2*delta__A+µ__A*µ__C*delta__A*omega__B+µ__A*µ__C*delta__A*sigma__B

 (36)
> 

NULL

"(->)"

Vector[column](%id = 18446746579340117046)

 (37)

# to find the trace we

 

Matrix(7, 7, {(1, 1) = -beta*lambda-v__1-µ, (1, 2) = v__2, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (1, 6) = 0, (1, 7) = 0, (2, 1) = v__1, (2, 2) = beta*(w-1)*lambda-µ-v__2-alpha, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (2, 6) = 0, (2, 7) = 0, (3, 1) = 0, (3, 2) = alpha, (3, 3) = -µ, (3, 4) = 0, (3, 5) = `ρ__A`, (3, 6) = `ρ__F`, (3, 7) = -(-1+k)*`ρ__Q`, (4, 1) = beta*lambda, (4, 2) = -beta*(w-1)*lambda, (4, 3) = 0, (4, 4) = -q__E-delta-µ, (4, 5) = 0, (4, 6) = 0, (4, 7) = 0, (5, 1) = 0, (5, 2) = 0, (5, 3) = 0, (5, 4) = a*delta, (5, 5) = -`ρ__A`-q__A-µ, (5, 6) = 0, (5, 7) = k*`ρ__Q`, (6, 1) = 0, (6, 2) = 0, (6, 3) = 0, (6, 4) = -delta*(-1+a), (6, 5) = 0, (6, 6) = -`ρ__F`-q__F-`δ__F`-µ, (6, 7) = 0, (7, 1) = 0, (7, 2) = 0, (7, 3) = 0, (7, 4) = q__E, (7, 5) = q__A, (7, 6) = q__F, (7, 7) = -`ρ__Q`-`δ__Q`-µ})

Matrix(%id = 36893490965935089652)

 (38)

"(->)"

-beta*lambda-v__1-7*µ+beta*(w-1)*lambda-v__2-alpha-q__E-delta-rho__A-q__A-rho__F-q__F-delta__F-rho__Q-delta__Q

 (39)

 

#this shows that trace is negative

 

#to Achieve stability, the value below must be less than zero

 

(-q__E-delta-µ)*(-`&rho;__F`-q__F-`&delta;__F`-µ)*(-k*q__A*`&rho;__Q`+q__A*µ+q__A*`&delta;__Q`+q__A*`&rho;__Q`+µ^2+µ*`&delta;__Q`+µ*`&rho;__A`+µ*`&rho;__Q`+`&delta;__Q`*`&rho;__A`+`&rho;__A`*`&rho;__Q`)*µ < 0

(-q__E-delta-µ)*(-rho__F-q__F-delta__F-µ)*(-k*q__A*rho__Q+q__A*rho__Q+q__A*µ+q__A*delta__Q+rho__A*rho__Q+rho__A*µ+rho__A*delta__Q+rho__Q*µ+µ^2+µ*delta__Q)*µ < 0

 (40)

 NULL

> 

M := diff(N(t), t) = Pi-µ*N(t)

diff(N(t), t) = Pi-µ*N(t)

 (41)
> 

dsolve({M}, N(t))

{N(t) = Pi/µ+exp(-µ*t)*_C1}

 (42)
> 

eval({N(t) = Pi/µ+exp(-µ*t)*_C1}, [t = infinity])

{N(infinity) = Pi/µ+exp(-µ*infinity)*_C1}

 (43)
> 

value(%)

{N(infinity) = Pi/µ+exp(-µ*infinity)*_C1}

 (44)
> 

Limit(N(t) = Pi/µ+exp(-µ*t)*_C1, t = infinity); value(%)

Limit(N(t) = Pi/µ+exp(-µ*t)*_C1, t = infinity)

  

limit(N(t), t = infinity) = limit(Pi/µ+exp(-µ*t)*_C1, t = infinity)

 (45)

 

> 

Subs := diff(S(t), t) = -(beta*lambda+v__1+µ)*S(t)

diff(S(t), t) = -(beta*lambda+v__1+µ)*S(t)

 (46)
> 

dsolve({Subs}, S(t))

{S(t) = _C1*exp(-(beta*lambda+v__1+µ)*t)}

 (47)
 

``

Download Cotton_DFE_and_Jacobian.mw

Help with Solving ODE and Representing ± Sign...

Asked by: salim-barzani 1555
ode dsolve differential-equations
April 29 2025
3  8

I tried solving this ODE, but my result is very different from the expected one. How can I correctly obtain the solution? Also, is there a way to include both the positive and negative signs (±) in the equation so that the final result reflects both possibilities?

> 

restart
> 

with(PDEtools)
> 

with(LinearAlgebra)
> 

with(Physics)
> 

with(SolveTools)
> 

undeclare(prime)

`There is no more prime differentiation variable; all derivatives will be displayed as indexed functions`

 (1)
> 

_local(gamma)

Warning, A new binding for the name `gamma` has been created. The global instance of this name is still accessible using the :- prefix, :-`gamma`.  See ?protect for details.

  
> 

declare(Omega(x, t)); declare(U(xi)); declare(u(x, y, z, t)); declare(Q(xi)); declare(V(xi))

Omega(x, t)*`will now be displayed as`*Omega

  

U(xi)*`will now be displayed as`*U

  

u(x, y, z, t)*`will now be displayed as`*u

  

Q(xi)*`will now be displayed as`*Q

  

V(xi)*`will now be displayed as`*V

 (2)
> 

``
> 

ode := f*g^3*(diff(diff(U(xi), xi), xi))-4*f*p*U(xi)-6*k*l*U(xi)-f^3*g*(diff(diff(U(xi), xi), xi))+6*f*g*U(xi)^2 = 0

f*g^3*(diff(diff(U(xi), xi), xi))-4*f*p*U(xi)-6*k*l*U(xi)-f^3*g*(diff(diff(U(xi), xi), xi))+6*f*g*U(xi)^2 = 0

 (3)
> 

S := (diff(G(xi), xi))^2-r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2) = 0

(diff(G(xi), xi))^2-r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2) = 0

 (4)
> 

S1 := dsolve(S, G(xi))

G(xi) = (1/2)*(-b+(-4*a*l+b^2)^(1/2))/l, G(xi) = -(1/2)*(b+(-4*a*l+b^2)^(1/2))/l, G(xi) = -4*a*exp(c__1*r*a^(1/2))/(exp(xi*r*a^(1/2))*(4*a*l-b^2+2*b*exp(c__1*r*a^(1/2))/exp(xi*r*a^(1/2))-(exp(c__1*r*a^(1/2)))^2/(exp(xi*r*a^(1/2)))^2)), G(xi) = -4*a*exp(xi*r*a^(1/2))/(exp(c__1*r*a^(1/2))*(4*a*l-b^2+2*b*exp(xi*r*a^(1/2))/exp(c__1*r*a^(1/2))-(exp(xi*r*a^(1/2)))^2/(exp(c__1*r*a^(1/2)))^2))

 (5)
> 

S2 := S1[3]

G(xi) = -4*a*exp(c__1*r*a^(1/2))/(exp(xi*r*a^(1/2))*(4*a*l-b^2+2*b*exp(c__1*r*a^(1/2))/exp(xi*r*a^(1/2))-(exp(c__1*r*a^(1/2)))^2/(exp(xi*r*a^(1/2)))^2))

 (6)
> 

normal(G(xi) = -4*a*exp(c__1*r*a^(1/2))/(exp(xi*r*a^(1/2))*(4*a*l-b^2+2*b*exp(c__1*r*a^(1/2))/exp(xi*r*a^(1/2))-(exp(c__1*r*a^(1/2)))^2/(exp(xi*r*a^(1/2)))^2)), ':-expanded')

G(xi) = 4*a*exp(c__1*r*a^(1/2))*exp(xi*r*a^(1/2))/(-4*a*l*(exp(xi*r*a^(1/2)))^2+b^2*(exp(xi*r*a^(1/2)))^2-2*b*exp(c__1*r*a^(1/2))*exp(xi*r*a^(1/2))+(exp(c__1*r*a^(1/2)))^2)

 (7)
> 

simplify(G(xi) = 4*a*exp(c__1*r*a^(1/2))*exp(xi*r*a^(1/2))/(-4*a*l*(exp(xi*r*a^(1/2)))^2+b^2*(exp(xi*r*a^(1/2)))^2-2*b*exp(c__1*r*a^(1/2))*exp(xi*r*a^(1/2))+(exp(c__1*r*a^(1/2)))^2))

G(xi) = -4*a*exp(a^(1/2)*r*(c__1+xi))/(4*a*l*exp(2*xi*r*a^(1/2))-b^2*exp(2*xi*r*a^(1/2))+2*b*exp(a^(1/2)*r*(c__1+xi))-exp(2*c__1*r*a^(1/2)))

 (8)
> 

convert(%, trig)

G(xi) = -4*a*(cosh(a^(1/2)*r*(c__1+xi))+sinh(a^(1/2)*r*(c__1+xi)))/(4*a*l*(cosh(2*xi*r*a^(1/2))+sinh(2*xi*r*a^(1/2)))-b^2*(cosh(2*xi*r*a^(1/2))+sinh(2*xi*r*a^(1/2)))+2*b*(cosh(a^(1/2)*r*(c__1+xi))+sinh(a^(1/2)*r*(c__1+xi)))-cosh(2*c__1*r*a^(1/2))-sinh(2*c__1*r*a^(1/2)))

 (9)
> 

convert(S1[3], trig)

G(xi) = -4*a*(cosh(c__1*r*a^(1/2))+sinh(c__1*r*a^(1/2)))/((cosh(xi*r*a^(1/2))+sinh(xi*r*a^(1/2)))*(4*a*l-b^2+2*b*(cosh(c__1*r*a^(1/2))+sinh(c__1*r*a^(1/2)))/(cosh(xi*r*a^(1/2))+sinh(xi*r*a^(1/2)))-(cosh(c__1*r*a^(1/2))+sinh(c__1*r*a^(1/2)))^2/(cosh(xi*r*a^(1/2))+sinh(xi*r*a^(1/2)))^2))

 (10)
> 

simplify(G(xi) = -4*a*(cosh(c__1*r*a^(1/2))+sinh(c__1*r*a^(1/2)))/((cosh(xi*r*a^(1/2))+sinh(xi*r*a^(1/2)))*(4*a*l-b^2+2*b*(cosh(c__1*r*a^(1/2))+sinh(c__1*r*a^(1/2)))/(cosh(xi*r*a^(1/2))+sinh(xi*r*a^(1/2)))-(cosh(c__1*r*a^(1/2))+sinh(c__1*r*a^(1/2)))^2/(cosh(xi*r*a^(1/2))+sinh(xi*r*a^(1/2)))^2)))

G(xi) = -4*a*(cosh(c__1*r*a^(1/2))+sinh(c__1*r*a^(1/2)))*(cosh(xi*r*a^(1/2))+sinh(xi*r*a^(1/2)))/((4*a*l-b^2)*cosh(xi*r*a^(1/2))^2+((8*a*l-2*b^2)*sinh(xi*r*a^(1/2))+2*b*(cosh(c__1*r*a^(1/2))+sinh(c__1*r*a^(1/2))))*cosh(xi*r*a^(1/2))+(4*a*l-b^2)*sinh(xi*r*a^(1/2))^2+2*b*(cosh(c__1*r*a^(1/2))+sinh(c__1*r*a^(1/2)))*sinh(xi*r*a^(1/2))-(cosh(c__1*r*a^(1/2))+sinh(c__1*r*a^(1/2)))^2)

 (11)
   

Download tt.mw

Slider Plot and Value Substitution Help...

Asked by: Andiguys 65
parameters explore plotting
April 29 2025
0  4

I need to create a slider plot for A10, A11, and A12 by varying the parameters theta, Pu, and a.
I have a syntax ready — could you suggest modifications to make it work correctly and generate the plot?

Additionally, is it possible to compute the values of A13 and A14 by substituting the obtained A10, A11, and A12 values for each combination of theta, Pu, and a from the slider plot?

Sheet attached: Slider_Q.mw

bug report. dsolve sometimes gives series solution...

Asked by: nm 11353
ode series differential-equations
April 28 2025
1  3

The series to ode using 'series' option (if it exists) should always be series(...), i.e. with big O at end. but sometimes Maple forgets to add this. Here is an example

> 

interface(version);

`Standard Worksheet Interface, Maple 2025.0, Linux, March 24 2025 Build ID 1909157`

> 

Physics:-Version();

`The "Physics Updates" version in the MapleCloud is 1862 and is the same as the version installed in this computer, created 2025, April 25, 10:33 hours Pacific Time.`

> 

SupportTools:-Version();

`The Customer Support Updates version in the MapleCloud is 15 and is the same as the version installed in this computer, created April 27, 2025, 23:18 hours Eastern Time.`

> 

restart;
> 

ode:=diff(y(x),x)+y(x)=1+x;
IC:=y(0)=0;
sol:=dsolve([ode,IC],y(x),'series')

diff(y(x), x)+y(x) = 1+x

y(0) = 0

y(x) = x

> 

lprint(sol); # notice solution is not series, it should be

y(x) = x

> 

#above solution should be
y(x) = series(x+O(x^6),x,6)

y(x) = series(x+O(x^6),x,6)

> 

#this example below is correct
ode:=diff(y(x),x)+y(x)=1+x;
IC:=y(0)=1;
sol:=dsolve([ode,IC],y(x),'series')

diff(y(x), x)+y(x) = 1+x

y(0) = 1

y(x) = series(1+(1/2)*x^2-(1/6)*x^3+(1/24)*x^4-(1/120)*x^5+O(x^6),x,6)

> 

lprint(sol); #solution is series

y(x) = series(1+1/2*x^2-1/6*x^3+1/24*x^4-1/120*x^5+O(x^6),x,6)

 

 

Download bug_report_dsolve_series_april_28_2025.mw

Controlling Variable Distribution in Polynomial Fo...

Asked by: salim-barzani 1555
polynomial expand pde
April 28 2025
2  5

In this work, I do not intend to expand all the variables across the monomials. Instead, I want to restrict the distribution to only the variables x,y,z,tx, y, z, tx,y,z,t, possibly raising them to appropriate powers as needed, until I obtain the desired solution and satisfy the conditions of my PDE tests. However, I am uncertain whether "monomial" is the correct term to use here.

S1.mw

trail-1.mw

why such different result from integrate when pull...

Asked by: nm 11353
integration int indefinite-integration
April 27 2025
2  3

These two expressions are the same, just pulled minus sign out

But look what happens when integrating them. the anti derivative of one is much more complicated than the other and contains complex numbers and logs. And no matter what I tried, I could not convert the complicated one to look same as the simpler result. Also could not verify the complicated one by back differentiating.

> 

integrand_1:=x^2*(-arctan(x) + x)*exp(-arctan(x) + x)/(x^2 + 1);

x^2*(-arctan(x)+x)*exp(-arctan(x)+x)/(x^2+1)

> 

integrand_2:=evala(integrand_1);

-x^2*(arctan(x)-x)*exp(-arctan(x)+x)/(x^2+1)

> 

simplify(integrand_1 - integrand_2)

0

> 

anti_1:=int(integrand_1,x);

(-arctan(x)+x)*exp(-arctan(x)+x)-exp(-arctan(x)+x)

> 

anti_2:=int(integrand_2,x);

-(1-x+((1/2)*I)*ln(1-I*x)-((1/2)*I)*ln(1+x*I))*(1-I*x)^(-(1/2)*I)*(1+x*I)^((1/2)*I)*exp(x)

> 

simplify(diff(anti_1,x)-integrand_1);

0

> 

simplify(diff(anti_2,x)-integrand_2);

Error, (in simpl/simpl/ReIm/sum) too many levels of recursion

> 

simplify(anti_1 - anti_2)

Error, (in simpl/simpl/ReIm/sum) too many levels of recursion

> 

simplify(anti_2);

(1/2)*(I*ln(1+x*I)-I*ln(1-I*x)+2*x-2)*(1-I*x)^(-(1/2)*I)*(1+x*I)^((1/2)*I)*exp(x)

> 

simplify(anti_2,ln);

(1/2)*(I*ln(1+x*I)-I*ln(1-I*x)+2*x-2)*(1-I*x)^(-(1/2)*I)*(1+x*I)^((1/2)*I)*exp(x)

 

 

Download int_strange_result_april_27_2025.mw

I would have expected same anti derivative to show.  To check, I used another software, and that one gave same anti-derivative for both integrands.

The questions I have: Why Maple gives such different result for same integrand? And how could one convert the one with the logs and complex numbers to the first one?

Maple 2025

Compute integral exactly ...

Asked by: Zeineb 540
integration
April 27 2025
0  0

Dear all 
I have a double integral, i want to compute this integral and verify if the pproposed solution verify the proposed equation or not. 
I can modify the right hand side of my equation or the exact solution, so that my equation has an exact solution with simple form of right hand side. 

exact_solution.mw

Thank you for your help 

Plotting contour integral diagrams...

Asked by: Paras31 245
integration contour plotting
April 27 2025
1  7

 I am writing notes on complex analysis, I need to use figures of contour paths to integrate on them, i want to create something like this

I tried to plot the contour for 
\oint_{|z|=2} \frac{1}{z^2+1}\,dz
I need to have connecting lines all around because the poles can not be isolated

> 

with(plots); circle1 := plot([2*cos(t), 2*sin(t), t = 0 .. 2*Pi], color = blue, thickness = 2); circle2 := plot([(1/2)*cos(t), 1+(1/2)*sin(t), t = 0 .. 2*Pi], color = "Green", thickness = 2); circle3 := plot([(1/2)*cos(t), -1+(1/2)*sin(t), t = 0 .. 2*Pi], color = "Red", thickness = 2); sing1 := plottools[disk]([0, 1], 0.2e-1, color = white); sing2 := plottools[disk]([0, -1], 0.2e-1, color = white); label1 := textplot([.1, 1.1, "z = i"], font = [Arial, Bold, 12]); label2 := textplot([.1, -1.1, "z = -i"], font = [Arial, Bold, 12])
> 

display(circle1, circle2, circle3, sing1, sing2, label1, label2, scaling = constrained, labels = ["Re", "Im"])

 
 
> 

restart; f := proc (z) options operator, arrow; 1/(z^2+1) end proc; z := 2*exp(I*t); dz := diff(z, t); integrand := f(z)*dz; simplify(integrand); value(Int(integrand, t = 0 .. 2*Pi))

0

 (1)

Download CIF.mw

about updating the Physics Package...

Asked by: Jean-Michel 40
update physics
April 27 2025
0  2

Hello all,

After updating the Physics package I have this error :

Physics:-Version();
The "Physics Updates" version "1862" is installed in the

   directory C:\Users\jm\maple\toolbox\2025\Physics Updates but

   is not active. The active version of Physics is within the

   library C:/Users/jm/maple/toolbox/2025/Physics Updates/lib\Ph\

  ysics Updates.maple.

What am I supposed to do next?

Thanks a lot and kind regards to all,

Jean-Michel

bug report. Maple 2025 server crash on call to ode...

Asked by: nm 11353
ode differential-equations crash
April 27 2025
2  1

FYI;

 

You might have to try the command more than one time to see the above crash. Here is the worksheet

> 

restart;
> 

interface(version);

`Standard Worksheet Interface, Maple 2025.0, Linux, March 24 2025 Build ID 1909157`

> 

Physics:-Version();

`The "Physics Updates" version in the MapleCloud is 1862 and is the same as the version installed in this computer, created 2025, April 25, 10:33 hours Pacific Time.`

> 

SupportTools:-Version();

`The Customer Support Updates version in the MapleCloud is 13 and is the same as the version installed in this computer, created April 22, 2025, 15:14 hours Eastern Time.`

> 

restart;
> 

ode:=x^2-2*x*y(x)+5*y(x)^2 = (x^2+2*x*y(x)+y(x)^2)*diff(y(x),x);

x^2-2*x*y(x)+5*y(x)^2 = (x^2+2*x*y(x)+y(x)^2)*(diff(y(x), x))

> 

sol:=y(x) = (-1/2*exp(RootOf(-exp(_Z)^2*ln(x*(exp(_Z)-2))+2*_C7*exp(_Z)^2+_Z*exp(_Z)^2+4*exp(_Z)*ln(x*(exp(_Z)-2))-8*_C7*exp(_Z)-4*exp(_Z)*_Z-2*exp(_Z)-4*ln(x*(exp(_Z)-2))+8*_C7+4*_Z+6))^2+3*exp(RootOf(-exp(_Z)^2*ln(x*(exp(_Z)-2))+2*_C7*exp(_Z)^2+_Z*exp(_Z)^2+4*exp(_Z)*ln(x*(exp(_Z)-2))-8*_C7*exp(_Z)-4*exp(_Z)*_Z-2*exp(_Z)-4*ln(x*(exp(_Z)-2))+8*_C7+4*_Z+6))-6+2*(exp(RootOf(-exp(_Z)^2*ln(x*(exp(_Z)-2))+2*_C7*exp(_Z)^2+_Z*exp(_Z)^2+4*exp(_Z)*ln(x*(exp(_Z)-2))-8*_C7*exp(_Z)-4*exp(_Z)*_Z-2*exp(_Z)-4*ln(x*(exp(_Z)-2))+8*_C7+4*_Z+6))^2-6*exp(RootOf(-exp(_Z)^2*ln(x*(exp(_Z)-2))+2*_C7*exp(_Z)^2+_Z*exp(_Z)^2+4*exp(_Z)*ln(x*(exp(_Z)-2))-8*_C7*exp(_Z)-4*exp(_Z)*_Z-2*exp(_Z)-4*ln(x*(exp(_Z)-2))+8*_C7+4*_Z+6))+9)^(1/2))/(1/2*exp(RootOf(-exp(_Z)^2*ln(x*(exp(_Z)-2))+2*_C7*exp(_Z)^2+_Z*exp(_Z)^2+4*exp(_Z)*ln(x*(exp(_Z)-2))-8*_C7*exp(_Z)-4*exp(_Z)*_Z-2*exp(_Z)-4*ln(x*(exp(_Z)-2))+8*_C7+4*_Z+6))^2-3*exp(RootOf(-exp(_Z)^2*ln(x*(exp(_Z)-2))+2*_C7*exp(_Z)^2+_Z*exp(_Z)^2+4*exp(_Z)*ln(x*(exp(_Z)-2))-8*_C7*exp(_Z)-4*exp(_Z)*_Z-2*exp(_Z)-4*ln(x*(exp(_Z)-2))+8*_C7+4*_Z+6)))*x:
> 

odetest(sol,ode);
> 

 

Download crash_maple_2025_april_27_2025.mw

Hopefully a fix could be found for this.

Initialization or Library Archive...

Asked by: Christopher2222 6030
April 27 2025
0  2

When generating a file to update parts of maple (for example constants) is it best to put it in an initialization file or make a library archive .mla?

ScientificConstants names ...

Asked by: Christopher2222 6030
subscript scientificconstants rtfm
April 27 2025
0  5

I fail to see the logic of using short form symbols for the scientific constants and then not being able to use that short form.  One manually has to equate the two as I show below.  Anyone see a reason not to do that in the internal programming?  Just wondering. 

with(ScientificConstants):

GetConstant(M__Sun)             

One has to use the names associated with those short forms described by the command

GetConstant(mass_of_Sun)

I would much rather like to use MSun , so what one has to do is manually equate them

MSun:=mass_of_Sun:

GetConstant(MSun)

Separating solutions based on PDE satisfaction...

Asked by: salim-barzani 1555
pde differential-equations select
April 27 2025
2  6

I have a list of candidate solutions. Some of them satisfy my PDE test (i.e., they make the PDE equal to zero), while others do not. How can I separate the solutions that satisfy the PDE from those that do not?

Trail-pdetest.mw

how do i solve a vector question...

Asked by: Hidayat 5
April 27 2025
1  2

ABC is an equilateral triangle of side 3 units. The points P, Q lie on BC, CA re-
spectively and are such that AQ = CP = 2units. If the point R lies on AB produced

so that BR = 1unit, prove that P, Q, R are collinear.

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