Parham2016

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7 years, 118 days

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These are questions asked by Parham2016

 

Hello all. Is there any solution for the indefinite integralBadIntegral.mw

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int((B*alpha[1]+(1/8)*`Νu`)*HeunT((1/16)*alpha[1]*(8*B*alpha[1]+`Νu`)*3^(2/3)*(2*alpha[2]+1)^2/(alpha[2]*(alpha[1]*alpha[2]*(8*B*alpha[1]+`Νu`))^(1/3)), 0, (1/2)*alpha[1]*(8*B*alpha[1]+`Νu`)*3^(1/3)/(alpha[1]*alpha[2]*(8*B*alpha[1]+`Νu`))^(2/3), (1/3)*3^(2/3)*(alpha[1]*alpha[2]*(8*B*alpha[1]+`Νu`))^(1/6)*y), y)

int((B*alpha[1]+(1/8)*`Νu`)*HeunT((1/16)*alpha[1]*(8*B*alpha[1]+`Νu`)*3^(2/3)*(2*alpha[2]+1)^2/(alpha[2]*(alpha[1]*alpha[2]*(8*B*alpha[1]+`Νu`))^(1/3)), 0, (1/2)*alpha[1]*(8*B*alpha[1]+`Νu`)*3^(1/3)/(alpha[1]*alpha[2]*(8*B*alpha[1]+`Νu`))^(2/3), (1/3)*3^(2/3)*(alpha[1]*alpha[2]*(8*B*alpha[1]+`Νu`))^(1/6)*y), y)

(1)

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Thanks

 

Hello all,

 

Is here anyone who could help me with this Integral.

I want to determine P(y=2) in which the Nu is a constant and also is unknown.

In deed I want to derive the P which only has one unknown parameter(Nu). But I cannot solve this indefinite integral although the \alpha_1 , \alpha_2 and B are known constants. The Nu parameter mustn't be valued preferely.Indefinite_INtegral.mw

 Hi all,

 Is there anyone who could help me with this error? I am sure there is at least one solution for the equation.

 Thanks

Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/EQ.mw .

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Hi all,

 

Please help me with this question. I want to solve a PDE by Maple.

restart

A := 5;

5

(1)

B := 9

9

(2)

c := 1

1

(3)

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eq := diff(u(x, t), t, t)-c^2*(diff(u(x, t), x, x));

diff(diff(u(x, t), t), t)-(diff(diff(u(x, t), x), x))

(4)

 

dsolve({eq, u(0, t) = A, u(1, t) = B, u(x, 0) = 0, (D(u))(x, 0) = sin(x)}, u(x, t))

Error, (in evalapply/D) too many variables for the derivative of a function of only one variable in D(u)(x, 0)

 

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Hi all,

 

It is believeble for me that the KummerU function is infinite or has a singular solution at r=0;

but without the command D(R) and dsolve ({er, D(R)(0)=0}, {R(r)}) that show the r=0 is a singular solution for the R(r), how could I prove it without using of these commands: for example is plotting of this function applicable??

while we have not the value of the  \lambda

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restart

eq := diff(R(r), r, r)+(diff(R(r), r))/r+(-r^2+1)*lambda^2*R(r) = 0;

diff(diff(R(r), r), r)+(diff(R(r), r))/r+(-r^2+1)*lambda^2*R(r) = 0

(1)

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dsolve(eq);

R(r) = _C1*exp(-(1/2)*lambda*r^2)*KummerM(1/2-(1/4)*lambda, 1, lambda*r^2)+_C2*exp(-(1/2)*lambda*r^2)*KummerU(1/2-(1/4)*lambda, 1, lambda*r^2)

(2)

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dsolve({eq, (D(R))(0) = 0}, {R(r)})

R(r) = _C1*exp(-(1/2)*lambda*r^2)*KummerM(1/2-(1/4)*lambda, 1, lambda*r^2)

(3)

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R := proc (r) options operator, arrow; C1*exp(-(1/2)*lambda*r^2)*KummerM(1/2-(1/4)*lambda, 1, lambda*r^2)+_C2*exp(-(1/2)*lambda*r^2)*KummerU(1/2-(1/4)*lambda, 1, lambda*r^2) end proc

proc (r) options operator, arrow; C1*exp(-(1/2)*lambda*r^2)*KummerM(1/2-(1/4)*lambda, 1, lambda*r^2)+_C2*exp(-(1/2)*lambda*r^2)*KummerU(1/2-(1/4)*lambda, 1, lambda*r^2) end proc

(4)

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D(R)

proc (r) options operator, arrow; -C1*lambda*r*exp(-(1/2)*lambda*r^2)*KummerM(1/2-(1/4)*lambda, 1, lambda*r^2)+2*C1*exp(-(1/2)*lambda*r^2)*((lambda*r^2-1/2-(1/4)*lambda)*KummerM(1/2-(1/4)*lambda, 1, lambda*r^2)+(1/2+(1/4)*lambda)*KummerM(-1/2-(1/4)*lambda, 1, lambda*r^2))/r-_C2*lambda*r*exp(-(1/2)*lambda*r^2)*KummerU(1/2-(1/4)*lambda, 1, lambda*r^2)+2*_C2*exp(-(1/2)*lambda*r^2)*((lambda*r^2-1/2-(1/4)*lambda)*KummerU(1/2-(1/4)*lambda, 1, lambda*r^2)-KummerU(-1/2-(1/4)*lambda, 1, lambda*r^2))/r end proc

(5)

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