2. Find the solution to the following non-homogeneous heat equation with given initial and boundary conditions on [0, π]. Ut - Uxx = e -2t sin (3x), u(0, t) = u(π, t) = 0, u(x,0) = = sin x.
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![2. Find the solution to the following non-homogeneous heat equation with given initial and boundary
conditions on [0, π].
Ut
Uxx = e
-2t sin (3x),
u(0, t) = u(π, t) = 0,
u(x,0) = sinx.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F91f76606-a4d9-42f0-be0d-7b1366a5593f%2Fed6210c4-05ba-425e-a935-6fc8c944164b%2Fh6gmbt_processed.png&w=3840&q=75)
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- For the IVP: (1-4) cos ty" -In(r-1)y +vT+5y=e, y(2) = 1, y'(2) = 1 determine the largest interval in which the solution is certain to exist B A (-5,4) (5,4). D (1,2 (1,00).Solve the heat equation U; = 4u xx, - 00, %3D u(x,0)=6(x-2). 8. O A Ux,1) == Je-dax cos[@x-2)]dm. -88 C8 OB U(x,f) = Se4at cos[@x-2)]d o. 1 Ocu(x,t) = Je-4 cos[@(x+2)]d o. -8 O D. None of these. 8. OE U(x,t) = 1 Te-Aat sin[ @x-2)]dw. %3D -48Apply the technique of separation of variables to partial differential equations and find the general solution for the homogeneous wave equation while considering periodic boundary conditions. Utt. = =&² Uxx, for x = (-a, a), t > 0 u(-a, t) = u(a, t), uz (-a, t) = u(a, t), for t > 0 U
- Consider the heat equation Ut = auxx, a € R⁰, 00 with boundary conditions ur(0,t) = ur(L,t)=0, t>0. i. Give a meaningful interpretation of the boundary conditions.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).Use the method of separation of variables and the superposition principle to solve the heat equation 192 subject to the boundary conditions (0, t) u(x, 0) = 0 0. and the initial condition 0, u(4, t) = 0, t > 0, Enter u(x, t). = 3 cos(³x) + cos(³), 01 ' The solution of the heat equation wzz =wt, 0Reduce following the heat conduction equation to the ordinary differential equations and separate the given homogeneous boundary conditions by using the method of separation of variables Ut=4ux u(0,t)=0, t>0 (the end x-0 is held at zero temperature) u,(2,t)=0, t>0 (no heat loss from this end) A X"-4AX=0, X(0)=0, X(2)=0 T'-AT=0, A0. X"-AX=0, X(0)-0, X(2)-0 T-4AT-0. A<0. X-AX-0, X(0)-0, X'(2)-0 T-4AT-0. A0. yazın7.(This one is a gift) Consider a homogeneous differential equation, p(x, y) dx + q(x, y) dy = 0, where the coefficients p(x, y) and q(x, y) satisfy the special homogeneity condition: p(Ax, Xy) = X®p(x, y); q(Ax, Ay) = X°q(x, y). Explain why it is always possible to express any such homogeneous differential equation in the form dy h. F dxSolve the one-dimensional heat equation (k = 1), du/dt = d2u/dx2 for a uniform rod of length 1 (ie, 0 < x < 1) subject to the boundary condtions: du/dx (0,t) = 0 du/dx (1,t) = 0 and with the initial condition: u(x,0) = 2cos(3πx) - 3 cos(5πx)Need full solution and add jot notes next to steps if needed so that I can better understand the solution.Please find the question attached.Recommended textbooks for youAdvanced Engineering MathematicsAdvanced MathISBN:9780470458365Author:Erwin KreyszigPublisher:Wiley, John & Sons, Incorporated
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