Question

Let u be the solution to the initial boundary value problem for the Heat Equation ut = 5u'' + 10x - 3 with Mixed boundary conditions u(0) = 0 and u(3) = 0 and with initial condition u(0,x) = 0 The solution u of the problem above, with the conventions given in class, has the form u(x,t) = c * e^(-5t) * sin(sqrt(10)t) * sin(sqrt(10)x) with the normalization conditions C = 1 and w = 1. Find the functions W and the constants c. W(x) = c * sin(sqrt(10)x) CA = 1

          Let u be the solution to the initial boundary value problem for the Heat Equation
ut = 5u'' + 10x - 3
with Mixed boundary conditions u(0) = 0 and u(3) = 0 and with initial condition
u(0,x) = 0

The solution u of the problem above, with the conventions given in class, has the form
u(x,t) = c * e^(-5t) * sin(sqrt(10)t) * sin(sqrt(10)x)

with the normalization conditions C = 1 and w = 1. Find the functions W and the constants c.

W(x) = c * sin(sqrt(10)x)
CA = 1
        
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let u be the solution to the initial boundary value problem for the heat equation utx5u1x0x03 with mixed boundary conditions ut00 and ut30 and with initial condition u0xx 0 the solution u of 98237

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Let u be the solution to the initial boundary value problem for the Heat Equation ut = 5u'' + 10x - 3 with Mixed boundary conditions u(0) = 0 and u(3) = 0 and with initial condition u(0,x) = 0 The solution u of the problem above, with the conventions given in class, has the form u(x,t) = c * e^(-5t) * sin(sqrt(10)t) * sin(sqrt(10)x) with the normalization conditions C = 1 and w = 1. Find the functions W and the constants c. W(x) = c * sin(sqrt(10)x) CA = 1
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Transcript

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00:01 Hello everyone here it is given that ut equal to 5uxx and t is greater than 0 and 0 less than x less than 1 so we can write that by separation of variables the most general solution is u of t comma x is equal to c1 cos px plus c2 sin px into e to the power minus 5p square t so uxt comma x will be minus c1 p sin px plus c2 p cos px e to the power minus 5p square but u of t comma 0 equal to 0 implies c1 e to the power minus 5p square t equal to 0 this implies c1 equal to 0 but ux of t comma 1 equal to 0 implies c2 p cos p equal to 0 so this implies cos px equal to 0 this in turn applies p equal to 2n minus 1 pi by 2 where n equal to 1 2 3 it goes on therefore by superposition…
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