#$&*
course Mth 279
3/26/14
Query 07 Differential Equations*********************************************
Question: 3.4.2. Solve the equation y ' = 2 t y ( 1 - y), with y(0) = -1, as a Bernoulli equation.
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Your solution:
->Bernoulli equation form: y' + p(t)y = q(t)y^n
y ' = 2 t y ( 1 - y)
y ' = 2 t y - 2ty^2
y ' - 2 t y = - 2ty^2
This equation is now in the correct form.
p(t) = -2t
q(t) = -2t
n = 2
Using v = y^m, which means y = v^(1/m), and m = (1 - n) from the Bernoulli equation.
The equation becomes:
dv/dt + (1-n)p(t)v = (1 - n)q(t)
dv/dt + (1-2)(-2t)v = (1 - 2)(-2t)
dv/dt + (-1)(-2t)v = (- 1)(-2t)
dv/dt + (2t)v = (2t)
This Equation is now First Order linear, and can be more easily solved
dv/dt + (2t)v = (2t)
dv/dt = (2t) - (2t)v
dv/dt = (2t) (1 - v)
dv/(1 - v) = 2t dt
integrate:
ln(1 - v) = t^2 + c
1 - v = Ce^(t^2)
not sure if I took the correct derivative
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Good so far.
y = v^m.
You need to solve the equation for y.
You also need to apply the initial conditions.
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):OK
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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:
->Bernoulli equation form: y' + p(t)y = q(t)y^n
y ' - y = t y^(1/3)
Since the Equation is already in the form:
p(t) = -1
q(t) = t
n = 1/3
Using v = y^m, —> 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)
@&
The right-hand side of the equation is just q(t).
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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)
use 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 yeilds:
ve^(-2t/3) = Int ((2/3)te^(-2t/3)) dt
Integration of the Right hand side
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), the Equation becomes:
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)
y(0) = (-0 - (3/2) + c)^(3/2)
(- (3/2) + c)^(3/2) = -9
(- (3/2) + c)^(3) = 81
(- (3/2) + c) = cube root (81)
C = cube root (81) + (3/2) = 5.83 approx.
Equation becomes:
y = (-t - (3/2) + cube root (81) + (3/2))^(3/2)
y = (-t + cube root (81))^(3/2)
y = (cube root (81) - t)^(3/2)
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):OK
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Self-critique rating:
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Question: 3.4.8. Solve the equation y ' = - (y + 1) + t ( y + 1)^(-2) as a Bernoulli equation.
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Your solution:
Is it possible to use substitution u=y+1?
@&
That is perfectly valid, and in this case it conveniently works out that du = dy.
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):Not sure on this one
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#*&!
&#Good responses. See my notes and let me know if you have questions. &#