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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.
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Your solution:
Proper Bernoulli form is an equation in form: y' + p(x)y = q(x)(y^n).
In order to get our equation into this form we will divide everything by y ( 1 - y):
y'/(y(1 - y)) = 2 t. We can tell from the equation that our parameter n is 0. Therefore, our substitution v = y^(1). Its derivative is v' = y
Substituting in v and v', we get v''/v'(1-v')) = 2t.
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary): The examples in the book and in the notes do not make any sense to me...there are no good online resources that make this example easier either...it seems like the examples are so much easier than these problems.
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Have you checked the course DVDs?
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Here's a solution to this problem:
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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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:
Proper Bernoulli form is an equation in form: y' + p(x)y = q(x)(y^n).
In order to get our equation into this form we will divide everything by y^(1/3):
y'/(y^(1/3)) - y/(y^(1/3)) = t. We can tell from the equation that our paramter n = 1/3. Therefore, our substitution v = y^(1/3). Its derivative is v' = 1/3y^(-2/3)y'
Substituting in v and v' we get y'/(v) - y/v = t
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:
Proper Bernoulli form is an equation in form: y' + p(x)y = q(x)(y^n).
In order to get our equation into this form we will add (y+1) and divide everything by ( y + 1)^(-2):
(y')/(( y + 1)^(-2)) + (y + 1)/(( y + 1)^(-2)) = t . We can tell from the equation that our paramter n = -2.
Therefore, our substitution v = y^(-2). Its derivative is v' = -2y^-3y'
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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Check my note, which includes a detailed solution to the first problem, then see if you can solve the other two equations.
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&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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