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course Mth 279
6/23 11:23
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 - 2ty^2
y' - 2ty = -2ty^2
n = 2, v = y^(1-2) = y^(-1)
v' - (-1)v = (-1) t
v' - v = -t
p(t) = -1
P(t) = -t
u = e^(-t)
e^(-t) v' - e(-t)v = -e^(-t)t
integrate
e^(-t) v = -te^(-t) - e^(-t) + C
v = -t -1 + Ce^(t)
-1 = 0 - 1 + Ce
0 = C
v = t-1
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We rearrange the equation to the form
y ' - 2 t y = 2 t y^2.
Letting v = y^m we get
v ' = m y^(m-1) * y '
so that
y ' = 1/m y^(1-m) v '
and since y = v^(1/m)
y ' = 1/m v^( (1 - m) / m) v '
Substituting for y and y ' we have
1/m v^( (1 - m) / m) v ' - 2 t v^(1/m) = 2 t v(2/m).
Multiplying both sides by m v^((1-m) / m) we get
v ' - 2 m t v = 2 m t v^(1/m + 1) .
The v on the right-hand side will 'disappear' if we let 1/m + 1 = 0, so that m = -1. The equation becomes
v ' + 2 t v = -2 t,
now a first-order linear homogeneous equation with integrating factor e^(t^2).
Multiplying by the integrating factor we have
e^(t^2) v ' + 2 t e^(t^2) v = 2 t e^(t^2).
The left-hand side is ( e^(t^2) v) ' so integration of the equation how yields
e^(t^2) v = integral(2 t e^(t^2) dt)
giving us
e^(t^2) v = e^(t^2) + c
v = e^(t^2 + c) / e^(t^2) = 1 + c e^(-t^2).
Since m = - 1, v = y^-1 so
1/y = 1 + c e^(-t^2)
and
y = 1 / (1 + c e^(-t^2)).
The initial condition y(0) = -1 gives us
-1 = 1 / (1 + c e^0) = 1 / (1 + c),
which is easily solved to obtain c = -2.
Thus
y = 1 / (1 - 2 e^(-t^2)).
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confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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Self-critique (if necessary):
I don't think I fully understand the purpose of this way of solving???
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Self-critique rating:
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Question: 3.4.6. Solve the equation y ' - y = t y^(1/3), y(0) = -9 as a Bernoulli equation.
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Your solution:
n = 1-1/3 = 2/3
v = y^(2/3)
v' - 2/3 = 2/3t
p(t) = -2/3
P(t) = -2/3 t
u = e^(-2/3 t)
e^(-2/3 t) v' - 2/3 e^(-2/3 t) = e^(-2/3 t) (2/3 t)
integrate
e^(-2/3 t) v = 2/3 (-2/3 te^(-2/3 t) - 4/9 e^(-2/3 t))C
v = -4/9 t - 8/27 + Ce^(-2/3t)
-9 = 0 -8/27 + C
C = -9 + 8/27 = -8.7037
v = -4/9 t - 8/27 - 8.7037e^(-2/3t)
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To solve y ' + p(t) y = q(t) y^n:
1. Let v = y^m, where m = 1 - n.
2. The resulting equation will be v ' + m p(t) v = q(t).
3. Solve for v.
4. Plug in y^m for v.
In this case, n = 1/3 so m = 1 - n = 1 - 1/3 = 2/3.
v = y^m = y^(2/3)
The resulting equation is
v ' - 2/3 v = t
This is solved using integrating factor e^(-2/3 t), obtaining
(v e^(-2/3 t) ) ' = t e^(-2/3 t)
Integrating both sides we get
v e^(-2/3 t) = -3/2 t e^(-2/3 t) - 9 / 4 e^(- 2/3 t) + c
so that
v = -3/2 t - 9/4 + c e^(2/3 t).
Since m = 2/3, v = y^(2/3) so
y^(2/3) = -3/2 t - 9/4 + c e^(2/3 t)
and
y = (-3/2 t - 9/4 + c e^(2/3 t))^(3/2)
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confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
I really don't think I should be getting fractions like 8/27
???
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Self-critique rating:
@&
There is nothing unusual about getting such fractions.
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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:
n = -2
v = y^(3)
v' = -(1+2) + (1+2) t v
v' = -3 + 3tv
v' - 3tv = -3
p(t) = -3t
P(t) = -3/2 t^2
u = e^(-3/2 t^2)
e^(-3/2 t^2) v' - 3t e^(-3/2 t^2) v = -3e^(-3/2 t^2)
integrate
e^(-3/2 t^2) v = -4/3 e^(-3/2 t^2) + C
v = -4/3 + Ce^(3/2 t^2) is the general solution
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
I'm still unsure why we solve this way
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Self-critique rating:
@&
For equations of this form, the method works.
No other known method does.
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