Assignment 9

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course Mth 279

6/25

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:

y’ = 2ty(1-y)

y’ = 2ty - 2ty^2

y’ - 2ty = -2ty^2

p(t) = -2t q(t) = -2t n = 2

v’ + (1 -n)p(t)v = (1-n)q(t)

v’ + (1 -2)(-2t)v = (1-2)(-2t)

v’ + 2tv = 2t

p(t) = 2t int(p(t)) = 2t^2/2 = t^2

e^t^2

e^(t^2)v’ + 2te^(t^2)v = 2te^(t^2)

integrate with respect to t on both sides

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

v = 1 + C/(e^(t^2))

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

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

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

-1 = 1/(1+Ce^(-0^2))

-(1+C) = 1

C = -2

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

confidence rating #$&*:

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1

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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:

p(t) = -1 q(t) = t n=(1/3)

v’ -(2/3)v = (2/3)t

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

e^((-2/3)t)

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

integrate by parts and then u-sub on right side of equation

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

(2/3)((-3/2)te^((-2/3)t) - int((-3/2)e^((-2/3)t) let w = (-2/3)t dw = -2/3

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

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

y = v^(3/2)

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

-9 = (0 - (3/2) + Ce^0)^3/2

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

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

C = (-9)^(2/3) + (3/2) = 5.83

y= (-t -(3/2) + 5.83e^((2/3)t))^(3/2)

confidence rating #$&*:

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1

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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:

Let z = y+1

therefore z’ = y’

z’ = -z + tz^-2

z’ + z = tz^-2

p(t) = 1 q(t)= t n=-2

v’ +3v = 3t

p(t) =3 int(p(t)) = 3t

e^(3t)v’ + 3e^(3t)v = 3e^(3t)t

integrate using by parts on right side of equation

e(3t)v =te^(3t) - (1/3)e^(3t) + C

v = t - (1/3) + Ce^(-3t)

z = v^(1/3)

z = (t - (1/3) + Ce^(-3t))^(1/3)

subbing back in z value

y + 1 = (t - (1/3) + Ce^(-3t))^(1/3)

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

confidence rating #$&*:

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2

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Given Solution:

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Self-critique (if necessary):

I don’t think I would have thought of letting z=y+1 without help.

Is there a different way of solving this problem???

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Self-critique rating:"

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Any other way of solving this problem would be much more difficult.

It's always important to look for possible changes of variable. When the change is linear, as in this case, it isn't that difficult to see what to try. In other cases it can be very challenging.

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Self-critique (if necessary):

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Self-critique (if necessary):

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&#Good responses. See my notes and let me know if you have questions. &#