query 7

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course Phy 2479

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Query 07 Differential Equations*********************************************

Question: 3.4.2. Solve the equation y ' = 2 t y ( 1 - y), with y(0) = -1, as a Bernoulli equation.

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Your solution:

change of dependent variable v = y^-1, v(0) = y(0)^-1 = -1

v’ - 1(-2tv) = -1(-2t)

v’ + 2tv = 2t, linear nonhomgenious mut thru by e^(Int(2t)) = e^(t^2)

e^(t^2)v’ + 2te^(t^2)v = 2te^(t^2)

e^(t^2)v = Int(2te^(t^2) dt ), u = t^2, du = 2t

Int(2te^(t^2)dt) = Int( e^(u)du) = e^u = e^(t^2) + C

e^(t^2)v = e^(t^2) + C

v = 1 + Ce^(-t^2)

For v(0) = y(0)^(-1) = -1

v = 1 + C = -1, C = -2

v(t) = 1 - 2e^(-t^2)

y = v^-1 = (1 - 2e^(-t^2))^-1 = 1/(1 - 2e^(-t^2)), will reduce further because of -t^2 but don’t have any time to waste.

y = 1/(1 - 2e^(-t^2))

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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Question: 3.4.6. Solve the equation y ' - y = t y^(1/3), y(0) = -9 as a Bernoulli equation.

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Your solution:

change of dependent variable v = y^(2/3), v(0) = y(0)^(2/3) = -9^(2/3)

v’ + (2/3)(-1)v = (2/3)t

v’ - (2/3)v = (2/3)t, linear nonhomgenious mut thru by e^(Int(-2/3)) = e^((-2/3)t)

e^((-2/3)t)v’ - (2/3)e^((-2/3)t)v = (2/3)t e^((-2/3)t)

e^((-2/3)t)v = Int((2/3)t e^((-2/3)t) dt ), int by parts

Int(((2/3)t e^((-2/3)t) dt) = (2/3)[ te^((-2/3)t) + (3/2)((-3/2)e^((-2/3)t))]

= (2/3)te^((-2/3)t) - (3/2)e^((-2/3)t) + C

e^((-2/3)t)v = (2/3)t e^((-2/3)t) - (3/2) e^((-2/3)t) + C

v = [(2/3)t - (3/2)] + Ce^((2/3)t)

For v(0) = y(0)^(2/3) = -9^(2/3)

v = -(3/2) + C = -9^(2/3), C = (3/2) - 9^(2/3) ????This doesn’t seem right for init

condition????

v(t) = [(2/3)t - (3/2)] + ((3/2) - 9^(2/3))e^((2/3)t)

y = v^(1/3)= (((2/3)t - (3/2)) + ((3/2) - 9^(2/3))e^((2/3)t))^(1/3)

y(t) = (((2/3)t - (3/2)) + ((3/2) - 9^(2/3))e^((2/3)t))^(1/3)