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Question:how I can find the solution of differential equations with dsolve?

Posted: NIMA112 35 Product: Maple 2022

November 09 2023

0

how I can find the solution of differential equations with dsolve?

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(1)

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(2)

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EQ1 := diff(U[1, n](x), x, x)-upsilon^2*U[1, n](x) = -2*(int(Q[1]*Dirac(x-x[1])*Dirac(eta-y[1])*sin(upsilon*eta), eta = 0 .. b))/b

diff(diff(U[1, 3](x), x), x)-(25/4)*U[1, 3](x) = 14.55468946*Dirac(x+1.)

(3)

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EQ2 := -(diff(U[2, n](x), x, x))-upsilon^2*U[2, n](x) = -2*(int(Q[2]*Dirac(x-x[2])*Dirac(eta-y[2])*sin(upsilon*eta), eta = 0 .. b))/b

-(diff(diff(U[2, 3](x), x), x))-(25/4)*U[2, 3](x) = 12.73535328*Dirac(x-1.)

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bc := limit(U[2, n](x), x = infinity) < infinity, alpha*(diff(U[1, n](-a), x))-beta*U[1, n](-a) = 0, U[1, n](0) = U[2, n](0), (D(U[1, n]))(0) = lambda*(D(U[2, n]))(0)

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{-2.5*U[1, 3](-Pi) = 0, diff(diff(U[1, 3](x), x), x)-(25/4)*U[1, 3](x) = 14.55468946*Dirac(x+1.), -(diff(diff(U[2, 3](x), x), x))-(25/4)*U[2, 3](x) = 12.73535328*Dirac(x-1.), U[1, 3](0) = U[2, 3](0), (D(U[1, 3]))(0) = 0.1e-1*(D(U[2, 3]))(0), limit(U[2, 3](x), x = infinity) < infinity}

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Error, (in dsolve/numeric/process_input) system must be entered as a set/list of expressions/equations

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U[1](x, y) = sum(U[1, n](x)*sin(upsilon*y), n = 1 .. infinity)

Error, (in sum) summation variable previously assigned, 2nd argument evaluates to 3 = 1 .. infinity

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U[2](x, y) = sum(U[2, n](x)*sin(upsilon*y), n = 1 .. infinity)

Error, (in sum) summation variable previously assigned, 2nd argument evaluates to 3 = 1 .. infinity

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Download Thesis_(1).mw

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