3. Solve the heat equationдиk-for 2 < x < 4, and t >0with the boundary conditionsu(2, t) = 0 and u(4, t) = 0, for t > 0,%3Dand the initial value conditionu(x, 0) = x for x E [2, 4].

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Answered: 3. Solve the heat equation ди k- for 2…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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3b
Let us consider the following problem for the heat equation ди = 0 in (0,1) × (0,+∞), ốt u(x,0) = u,(x),Vx E (0,1), ди (0,t): (1) ди -(1,t) = 0, Vt > 0. Use the method of separation of variables to compute the solution of problem (1) if 1) и,(х) %3D1, Vx € (0,1). 2) u,(x)= cos77x, Vx e (0,1). 3) u,(x) = 3+2 cos7x,Vx € (0,1). 4) u,(x) = sin x, Vx e (0,1).
Determine a solution of the IVP
Find the solution to the initial-boundary value problem: ди 0 0, д (0, t) — (1, t) %3D0, и(х, 0) — 2л, Ox² , ди (1.t) — 0, ди t>0, dx 0 < x < 1. ||
Let y=v,(x) y,+ v,(x) y; be the solution of y"+p(x) y+q(x)y3 In(x), which obtained by the variation of parameters method. If y, ()=x, and y,(x)=17 find v,(x). 33 vzK) =7 x² ( - - Intx)) 17 Vz(x) = x +In(x*) 17 None 35 v,(x) = x² (Inx -) Gear my choice
Show that x(t) = aje+njt + aze- , jt.is equal to Asin(@n+P). %3D 3. Using the general solution: x(t) = A1cos(@nt)+A2sin(@nt) siculate the values of A1 , A2 in terms of their initial conditions x0, v0
Given yı (t) = t and y2(t) = t satisfy the corresponding homogeneous equation of t'y" – 2y = 1 – 2t°, t > 0 Then the general solution to the nonhomogeneous equation can be written as y(t) = c1y1(t) + ©2Y2(t) + Yp(t). Use variation of parameters to find a particular solution yp(t). Y,(t) = Tip: Before you use the formula Y29(t) Yı9(t) JW(n, y2) Yp = - Y1 + y2 W (y1, Y2) the ODE should be in the form y'' + a(t)y' + b(t)y = g(t)
i = -I + y – y' ý = - – y + 2ry (1) (2) Find a Lyapunov Function Candidate of the form V = x9 + dy", choose d,p,g appropriately.
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3. Solve the heat equation ди k- for 2 < x < 4, and t > 0 with the boundary conditions u(2, t) = 0 and u(4, t) = 0, for t > 0, %3D and the initial value condition u(x, 0) = x for x E [2, 4|.