Query 07

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course Mth 279

7/1 12

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

multiply 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))

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

confidence rating #$&*:

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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:

y' - y = t*y^(1/3)

Since the Equation is already in the bernoulli form:

p(t) = -1

q(t) = t

n = 1/3

Using v = y^m, which means y = v^(1/m), and m = (1 - n) from the Bernoulli equation.

The equation becomes:

dv/dt + (1-n)p(t)v = (1 - n)q(t)

dv/dt + (1-(1/3))(-1)v = (1 - (1/3))(t)

dv/dt + (2/3)(-1)v = (2/3)(t)

dv/dt + (- 2/3)v = (2/3)(t)

This Equation can now be solved using a previous method.

u(t) = e^(int(p(t))dt) for the form y' + p(t)y = g(t)

For this equation:

p(t) = (-2/3)

int(p(t)) dt = (-2/3)t = -2t/3

u(t) = e^(-2t/3)

v' + (- 2/3)v = (2/3)(t)

v' e^(-2t/3) + (- 2/3)ve^(-2t/3) = (2/3)(t)e^(-2t/3)

v' e^(-2t/3) - (2/3)ve^(-2t/3) = (2/3)(t)e^(-2t/3)

(ve^(-2t/3))' = (2/3)te^(-2t/3)

Integration of both sides yields

ve^(-2t/3) = Int ((2/3)te^(-2t/3)) dt

Integration of the Right hand side required Integration by parts

udv = uv - Int(vdu)

u = t

dv = e^(-2t/3) dt

du = dt

v = Int(dv) = (-3/2)e^(-2t/3)

The Equation becomes:

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

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

ve^(-2t/3) = -te^(-2t/3) - (3/2)e^(-2t/3)

dividing out u(t) = e^(-2t/3)

v = -t - (3/2) + c

y = v^(1/m) = v^(1/(1 - n))

y = v^(1/(1 - 1/3)) = v^(3/2)

y = (-t - (3/2) + c)^(3/2)

solve for C with initial condition y(0) = -9

y(0) = (-0 - (3/2) + c)^(3/2)

(-(3/2) + c)^(3/2) = -9

(-(3/2) + c)^(3) = 81

(-(3/2) + c) = cuberoot(81)

C = cuberoot(81) + (3/2) = 5.83 approx.

Equation becomes:

y = (-t - (3/2) + cuberoot(81) + (3/2))^(3/2)

y = (-t + cuberoot(81))^(3/2)

y = (cuberoot(81) - t)^(3/2)

confidence rating #$&*:

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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: 3.4.8. Solve the equation y ' = - (y + 1) + t ( y + 1)^(-2) as a Bernoulli equation.

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Your solution:

y ' = -(y + 1) + t( y + 1)^(-2)y' + y + 1

= t( y + 1)^(-2) y’ + y

= t( y + 1)^(-2) - 1

substitute u = y + 1 so that u' = y'

u ' = -u + t u^2

u ' + u = t*u^2.

This is now a bernoulli equation

n = 2.

v = u^m = u^(1 - n) = u^(-1).

The equation becomes

v ' - v = t.

integrating factor u = e^(-t)

(v e^(-t)) ' = t e^(-t).

Integrating both sides yields

v e^(-t) = -t*e^(-t) - e^(-t) + c

v = -t - 1 + c e^t,

u^(-1) = (-t - 1 + c e^t)

u = (-t - 1 + c e^t)^(-1) = 1 / (-t - 1 + c e^t)

u = y + 1

y + 1 = 1 / (-t - 1 + c e^t)

y = y + 1 = 1 / (-t - 1 + c e^t) - 1

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:"

Self-critique (if necessary):

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Self-critique rating:

Self-critique (if necessary):

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Self-critique rating:

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&#Very good responses. Let me know if you have questions. &#