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Material Type: Assignment; Professor: Fulling; Class: ADV ENGINEERING MATH; Subject: MATHEMATICS; University: Texas A&M University; Term: Unknown 1989;
Typology: Assignments
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(a) y′′^ + y′^ +
y x + 1
(b) y′′^ − y′^ +
y x + 1
In both cases [(a) and (b)] consider
0 < 1 , 0 < x < 1 , y(0) = 0, y(1) = 3.
(a) y′′^ + xy′^ + y = 0,
0 < 1 , 2 < x < 4 , y(2) = 0, y(4) = 1.
(b) y′′^ − xy′^ + y = 0,
0 < 1 , 2 < x < 4 , y(2) = 0, y(4) = 1.
(c) y′′^ + xy′^ + y = 0,
0 < 1 , − 4 < x < − 2 , y(−4) = 1, y(−2) = 0.
Do not work out (c) from the beginning; knowing the answers to (a) and (b), you can solve (c) in two or three lines by a change of variable.
y
0 < 1 , 0 < x < 1 , y(0) = 2, y(1) = 1.
0 < 1 , 0 < x < 1 , y(0) = 1, y(1) = − ln 2.
0 < 1 , y(0) = y(1) = 1.
0 < 1 , y(0) = y(1) = 1.
Hint: You may find some help with the integral in a handbook under “Gudermannian function” or “Lobachevsky’s angle of parallelism”.
(a) utt − uxx = cos(x − t) (b) ut + u^2 ux = 0 (c) ut + 3t^2 u = 0 (d) utt − uxx = −m^2 u
∂u ∂x
(t, 0) = F 1 ,
∂u ∂x
(t, 1) = F 2.
(That is, the heat flux through each end of the bars is held constant.) What happens when you attempt to find a steady-state solution as on p. 66? Distinguish between the two cases F 1 = F 2 and F 1 6 = F 2. Can you give a physical explanation for your results?