Solve the heat equation Ou/dt = ko2u/dx2, 0 0 (L,t) = 0 t>0. (a)...

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Solve the heat equation Ou/dt = ko2u/dx2, 0<x<L, t> 0, subject to ди (0, 1) = 0 t>0 θα (L,t) = 0 t>0. дл (a) u(x, 0) = { 0 <L/2 Зла (b) u(x, 0)=6+4 cos 1 x > L/2 L 87x (c) u(x, 0) = -2sin (d) u(x, 0)=-3 cos Solve the heat equation Ou/dt = ko2u/dx2, 0<x<L, t> 0, subject to ди (0, 1) = 0 t>0 θα (L,t) = 0 t>0. дл (a) u(x, 0) = { 0 <L/2 Зла (b) u(x, 0)=6+4 cos 1 x > L/2 L 87x (c) u(x, 0) = -2sin (d) u(x, 0)=-3 cos

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To solve the heat equation with the given boundary and initial conditions we can use the method of separation of variables Lets denote the solution as ... View the full answer