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1. Find f(x, y) obeying = - sin(x)y and f(0, y) = }(y +2)². %3D %3D

1. Find f(x, y) obeying = - sin(x)y and f(0, y) = }(y +2)². %3D %3D

Advanced Engineering Mathematics
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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Question
1. Find f(x, y) obeying a² £_ Əzðy - sin(x)y and f(0, y) = }(y + 2)². 2. Solve for g(x, y, z) obeying Set 2, eq. (1) 3. Find a solution of = x² – 2 cos(t) with initial condition u(x,0) = cos(x) 4. Find a solution of u = x2 -2 cos(t) with initial condition u(x,0) = cos(x) %3D 5. In four lines or fewer, explain why af -ry se =xy, dx ду does not have a valid solution. 6. The heat equation, du = D dx² tends to “smooth out" kinks, so that its long term behaviour is given by, 0 = D Əx² Show that this results in a striaight line solution. 7. The wave equation, Pu Pu has a family of solutions u = f(x – vt) + g(x + vt) where ƒ can be any function. (a) Find f and g when u(x,0) = e=z*/2 and ®u(z.0) = e-z². */2; (b) Write down the solution u(x, t) when u(x,0) = e° (c) Plot (on the same graph) the solution fort=0, t = 1 and t = 5 8. Explain as simply as possible why u(x, t) = f(r – vt) for cannot be a valid solution for = -v if we have boundary conditions u(0, t) = u(L,t) = O with L > 0 and initial condition u(x,0) = g(x) # 0.
Transcribed Image Text:1. Find f(x, y) obeying a² £_ Əzðy - sin(x)y and f(0, y) = }(y + 2)². 2. Solve for g(x, y, z) obeying Set 2, eq. (1) 3. Find a solution of = x² – 2 cos(t) with initial condition u(x,0) = cos(x) 4. Find a solution of u = x2 -2 cos(t) with initial condition u(x,0) = cos(x) %3D 5. In four lines or fewer, explain why af -ry se =xy, dx ду does not have a valid solution. 6. The heat equation, du = D dx² tends to “smooth out" kinks, so that its long term behaviour is given by, 0 = D Əx² Show that this results in a striaight line solution. 7. The wave equation, Pu Pu has a family of solutions u = f(x – vt) + g(x + vt) where ƒ can be any function. (a) Find f and g when u(x,0) = e=z*/2 and ®u(z.0) = e-z². */2; (b) Write down the solution u(x, t) when u(x,0) = e° (c) Plot (on the same graph) the solution fort=0, t = 1 and t = 5 8. Explain as simply as possible why u(x, t) = f(r – vt) for cannot be a valid solution for = -v if we have boundary conditions u(0, t) = u(L,t) = O with L > 0 and initial condition u(x,0) = g(x) # 0.
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