Query 07

#$&*

course MTH 279

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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y’ = 2ty-2ty^2 y’ - 2ty = -2ty^2 y^(-2)y’ - 2ty^(-1) = -2t >> This in Bernoulli form with p(t) = -2t and q(t) = -2t v(t) = y^(-1) δv/δt = -y^(-2)*y’ -v’ - 2tv = -2t -v’ = -2t + 2tv v’ = 2t - 2tv v’ = 2t(1-v) ʃ1/(1-v) dv = ʃ2tdt -ln(v-1) + c_1 = t^2 + c_2 -ln(v-1) = t^2 + C ln(v-1) = -t^2 + C v-1 = Ae^(-t^2) v = Ae^(-t^2) + 1 v = y^(-1) >> Ae^(-t^2) + 1 = 1/y y = Ae^(t^2)/(Ae^(t^2) + 1) Imposing initial condition y(0) = -1, we get that A = -1/2. Substituting this in, we get: y(t) = e^(t^2)/(e^(t^2)-2) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: 3.4.6. Solve the equation y ' - y = t y^(1/3), y(0) = -9 as a Bernoulli equation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y’*y^(-1/3) - y^(2/3) = t v = y^(2/3) δv/δt = 2/3*y^(-1/3)*y’ 3/2v’ - v = t v’ - 2/3v = 2/3t µ(t) = e^(ʃ-2/3dt) = e^(-2t/3) e^(-2t/3)v’ - 2/3e^(-2t/3)v = 2/3te^(-2t/3) ʃd/dt(e^(-2t/3)v)dt = ʃ2/3*t*e^(-2t/3)dt v = (-2t-3)/2 + Ae^(2t/3) v = y^(2/3) (-2t-3)/2 + Ae^(2t/3) = y^(2/3) y = [(-2Ae^(2t/3)+2t+3)*sqrt(2(2Ae^(2t/3)-2t-3))]/4 Imposing initial condition y(0) = -9 A = [2(3^(4/3)+3)]/2 y(t) = -[((3(2(3^(1/3)) + 1)e^(2t/3) - 2t - 3)*sqrt(2(2(3^(1/3))+1)e^(2t/3)-2t-3)]/4 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I feel pretty confident that I have the right answer. It was a bit messy, and I hope I haven’t let any sign changes go astray. The final answer was a mess, so hopefully no parentheses are missing or incorrectly placed. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: 3.4.8. Solve the equation y ' = - (y + 1) + t ( y + 1)^(-2) as a Bernoulli equation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y’ = -y - 1 + t(y+1)^(-2) y’ + y = -1 + t (y+1)^(-2) x(t) = y(t) + 1 y = x-1 y’ = x’ x’ + (x-1) = -1 + t(x)^(-2) x^2*x’ + x^3 = t >> p(t) = 1, q(t) = t, n = -2 v = x^3 δv/δt = 3x^2*x’ v’/3 = x^2*x’ v’/3 + v = t v’ + 3v = 3t µ = e^(3t) e^(3t)v’ + 3e^(3t)v = 3te^(3t) ʃd/dt(e^(3t)v)dt = ʃ3te^(3t)dt e^(3t)*v = te^(3t) - e^(3t)/3 + C v = t - 1/3 + Ae^(-3t) x^3 = t - 1/3 + Ae^(-3t) x(t) = (t-1/3+Ae^(-3t))^(1/3) y(t) + 1 = (t - 1/3 + Ae^(-3t))^(1/3) y(t) = (t - 1/3 + Ae^(-3t))^(1/3) - 1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I spent a good bit of time on this problem, and I’m pretty sure that I tackled it the right way. I think what I have is correct. If there’s an easier/shorter way to do this problem, any advisement would be gladly appreciated. ------------------------------------------------ Self-critique rating: 3"

`gr31

Query 07

#$&*

course MTH 279

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y’ = 2ty-2ty^2

y’ - 2ty = -2ty^2

y^(-2)y’ - 2ty^(-1) = -2t >> This in Bernoulli form with p(t) = -2t and q(t) = -2t

v(t) = y^(-1)

δv/δt = -y^(-2)*y’

-v’ - 2tv = -2t

-v’ = -2t + 2tv

v’ = 2t - 2tv

v’ = 2t(1-v)

ʃ1/(1-v) dv = ʃ2tdt

-ln(v-1) + c_1 = t^2 + c_2

-ln(v-1) = t^2 + C

ln(v-1) = -t^2 + C

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

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

v = y^(-1) >> Ae^(-t^2) + 1 = 1/y

y = Ae^(t^2)/(Ae^(t^2) + 1)

Imposing initial condition y(0) = -1, we get that A = -1/2. Substituting this in, we get:

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

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

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Self-critique rating: OK

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y’*y^(-1/3) - y^(2/3) = t

v = y^(2/3)

δv/δt = 2/3*y^(-1/3)*y’

3/2v’ - v = t

v’ - 2/3v = 2/3t

µ(t) = e^(ʃ-2/3dt) = e^(-2t/3)

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

ʃd/dt(e^(-2t/3)v)dt = ʃ2/3*t*e^(-2t/3)dt

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

v = y^(2/3)

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

y = [(-2Ae^(2t/3)+2t+3)*sqrt(2(2Ae^(2t/3)-2t-3))]/4

Imposing initial condition y(0) = -9

A = [2(3^(4/3)+3)]/2

y(t) = -[((3(2(3^(1/3)) + 1)e^(2t/3) - 2t - 3)*sqrt(2(2(3^(1/3))+1)e^(2t/3)-2t-3)]/4

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I feel pretty confident that I have the right answer. It was a bit messy, and I hope I haven’t let any sign changes go astray. The final answer was a mess, so hopefully no parentheses are missing or incorrectly placed.

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Self-critique rating: 3

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y’ = -y - 1 + t(y+1)^(-2)

y’ + y = -1 + t (y+1)^(-2)

x(t) = y(t) + 1

y = x-1

y’ = x’

x’ + (x-1) = -1 + t(x)^(-2)

x^2*x’ + x^3 = t >> p(t) = 1, q(t) = t, n = -2

v = x^3

δv/δt = 3x^2*x’

v’/3 = x^2*x’

v’/3 + v = t

v’ + 3v = 3t

µ = e^(3t)

e^(3t)v’ + 3e^(3t)v = 3te^(3t)

ʃd/dt(e^(3t)v)dt = ʃ3te^(3t)dt

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

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

x^3 = t - 1/3 + Ae^(-3t)

x(t) = (t-1/3+Ae^(-3t))^(1/3)

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

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

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I spent a good bit of time on this problem, and I’m pretty sure that I tackled it the right way. I think what I have is correct. If there’s an easier/shorter way to do this problem, any advisement would be gladly appreciated.

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Self-critique rating: 3"

&#Good responses. Let me know if you have questions. &#