Question

Find the general solution u(t, x) of the boundary value problem for the heat equation with homogeneous Dirichlet boundary conditions ?_x u(t, 0) = 0, ?_x u(t, L) = 0. The c_n below denote arbitrary constants. u(t, x) = ?_{n=1}^{?} c_n e^{-k(frac{npi}{L})t} sin(frac{npi x}{L}) None of the options displayed. u(t, x) = frac{c_0}{2} + ?_{n=1}^{?} c_n e^{-k(frac{npi}{L})t} cos(frac{npi x}{L}) u(t, x) = frac{c_0}{2} + ?_{n=1}^{?} c_n e^{-k(frac{npi}{L})^2 t} cos(frac{npi x}{L}) u(t, x) = ?_{n=1}^{?} c_n e^{-k(frac{npi}{L})^2 t} sin(frac{npi x}{L}) u(t, x) = ?_{n=1}^{?} c_n e^{-k(frac{(2n-1)pi}{2L})^2 t} cos(frac{(2n-1)pi x}{2L}) u(t, x) = ?_{n=1}^{?} c_n e^{-k(frac{(2n-1)pi}{2L})^2 t} sin(frac{(2n-1)pi x}{2L}) We cannot find u(t, x) without an initial condition.

          Find the general solution u(t, x) of the boundary value problem for the heat equation with homogeneous Dirichlet boundary conditions
?_x u(t, 0) = 0, ?_x u(t, L) = 0.
The c_n below denote arbitrary constants.

u(t, x) = ?_{n=1}^{?} c_n e^{-k(frac{npi}{L})t} sin(frac{npi x}{L})
None of the options displayed.
u(t, x) = frac{c_0}{2} + ?_{n=1}^{?} c_n e^{-k(frac{npi}{L})t} cos(frac{npi x}{L})
u(t, x) = frac{c_0}{2} + ?_{n=1}^{?} c_n e^{-k(frac{npi}{L})^2 t} cos(frac{npi x}{L})
u(t, x) = ?_{n=1}^{?} c_n e^{-k(frac{npi}{L})^2 t} sin(frac{npi x}{L})
u(t, x) = ?_{n=1}^{?} c_n e^{-k(frac{(2n-1)pi}{2L})^2 t} cos(frac{(2n-1)pi x}{2L})
u(t, x) = ?_{n=1}^{?} c_n e^{-k(frac{(2n-1)pi}{2L})^2 t} sin(frac{(2n-1)pi x}{2L})
We cannot find u(t, x) without an initial condition.

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$Find the general solution u(t, x) of the boundary value problem for the heat equation with homogeneous Dirichlet boundary conditions ?x u(t, 0) = 0, ?x u(t, L) = 0. The cn below denote arbitrary constants. u(t, x) = ?n=1^? cn e^-k(fracnpiL)t sin(fracnpi xL) None of the options displayed. u(t, x) = fracc02 + ?n=1^? cn e^-k(fracnpiL)t cos(fracnpi xL) u(t, x) = fracc02 + ?n=1^? cn e^-k(fracnpiL)^2 t cos(fracnpi xL) u(t, x) = ?n=1^? cn e^-k(fracnpiL)^2 t sin(fracnpi xL) u(t, x) = ?n=1^? cn e^-k(frac(2n-1)pi2L)^2 t cos(frac(2n-1)pi x2L) u(t, x) = ?n=1^? cn e^-k(frac(2n-1)pi2L)^2 t sin(frac(2n-1)pi x2L) We cannot find u(t, x) without an initial condition.$

Added by Maxwell H.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Find the general solution u(t, x) of the boundary value problem for the heat equation with homogeneous Dirichlet boundary conditions ∂_x u(t,0) = 0, ∂_x u(t,L) = 0. The c_n below denote arbitrary constants.

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