#$&*
course Phy 2479
3/6 3
Query 07 Differential Equations*********************************************
Question: 3.4.2. Solve the equation y ' = 2 t y ( 1 - y), with y(0) = -1, as a Bernoulli equation.
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Your solution:
change of dependent variable v = y^-1, v(0) = y(0)^-1 = -1
v’ - 1(-2tv) = -1(-2t)
v’ + 2tv = 2t, linear nonhomgenious mut thru by e^(Int(2t)) = e^(t^2)
e^(t^2)v’ + 2te^(t^2)v = 2te^(t^2)
e^(t^2)v = Int(2te^(t^2) dt ), u = t^2, du = 2t
Int(2te^(t^2)dt) = Int( e^(u)du) = e^u = e^(t^2) + C
e^(t^2)v = e^(t^2) + C
v = 1 + Ce^(-t^2)
For v(0) = y(0)^(-1) = -1
v = 1 + C = -1, C = -2
v(t) = 1 - 2e^(-t^2)
y = v^-1 = (1 - 2e^(-t^2))^-1 = 1/(1 - 2e^(-t^2)), will reduce further because of -t^2 but don’t have any time to waste.
y = 1/(1 - 2e^(-t^2))
@& Good.
For reference:
We rearrange the equation to the form
y ' - 2 t y = 2 t y^2.
Letting v = y^m we get
v ' = m y^(m-1) * y '
so that
y ' = 1/m y^(m-1) v '
and since y = v^(1/m)
y ' = 1/m v^( (m - 1) / m).
Substituting for y and y ' we have
1/m v^( (m-1) / m) v ' - 2 t v^(1/m) = 2 t v(2/m).
Multiplying both sides by m v^((1-m) / m) we get
v ' - 2 m t v = 2 m t v^(1/m + 1) .
The v on the right-hand side will 'disappear' if we let 1/m + 1 = 0, so that m = -1. The equation becomes
v ' + 2 t v = -2 t,
now a first-order linear homogeneous equation with integrating factor e^(t^2).
Multiplying by the integrating factor we have
e^(t^2) v ' + 2 t e^(t^2) v = 2 t e^(t^2).
The left-hand side is ( e^(t^2) v) ' so integration of the equation how yields
e^(t^2) v = integral(2 t e^(t^2) dt)
giving us
e^(t^2) v = e^(t^2) + c
v = e^(t^2 + c) / e^(t^2) = 1 + c e^(-t^2).
Since m = - 1, v = y^-1 so
1/y = 1 + c e^(-t^2)
and
y = 1 / (1 + c e^(-t^2)).
The initial condition y(0) = -1 gives us
-1 = 1 / (1 + c e^0) = 1 / (1 + c),
which is easily solved to obtain c = -2.
Thus
y = 1 / (1 - 2 e^(-t^2)).
*@
confidence rating #$&*:
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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: 3.4.6. Solve the equation y ' - y = t y^(1/3), y(0) = -9 as a Bernoulli equation.
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Your solution:
change of dependent variable v = y^(2/3), v(0) = y(0)^(2/3) = -9^(2/3)
v’ + (2/3)(-1)v = (2/3)t
v’ - (2/3)v = (2/3)t, linear nonhomgenious mut thru by e^(Int(-2/3)) = e^((-2/3)t)
e^((-2/3)t)v’ - (2/3)e^((-2/3)t)v = (2/3)t e^((-2/3)t)
e^((-2/3)t)v = Int((2/3)t e^((-2/3)t) dt ), int by parts
Int(((2/3)t e^((-2/3)t) dt) = (2/3)[ te^((-2/3)t) + (3/2)((-3/2)e^((-2/3)t))]
= (2/3)te^((-2/3)t) - (3/2)e^((-2/3)t) + C
@& 3/2 * 3/2 - 9/4*@
e^((-2/3)t)v = (2/3)t e^((-2/3)t) - (3/2) e^((-2/3)t) + C
v = [(2/3)t - (3/2)] + Ce^((2/3)t)
For v(0) = y(0)^(2/3) = -9^(2/3)
v = -(3/2) + C = -9^(2/3), C = (3/2) - 9^(2/3) ????This doesn’t seem right for init
condition????
v(t) = [(2/3)t - (3/2)] + ((3/2) - 9^(2/3))e^((2/3)t)
y = v^(1/3)= (((2/3)t - (3/2)) + ((3/2) - 9^(2/3))e^((2/3)t))^(1/3)
y(t) = (((2/3)t - (3/2)) + ((3/2) - 9^(2/3))e^((2/3)t))^(1/3)
@& 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 ' + p(t) v^(1/m) = q(t) v^(n / m).
Multiplying both sides by m v^( (m - 1) / m) we have
v ' + m p(t) v = q(t) v^(( n + m - 1) / m)
Now, if the v thing on the right-hand side would just go away, the equation would be first-order linear.
We can arrange that. If the exponent (n + m - 1) is zero, then v^((n + m - 1) / m) will be v^0 = 1.
Remember that we haven't specified the value of m. It can be anything.
If we choose m so that n + m - 1 = 0, which requires that m = 1 - n, we will have our linear equation.
So, in a nutshell:
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)*@
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:
y ' = - (y + 1) + t ( y + 1)^(-2) y ' + y + 1 = t ( y + 1)^(-2) y’ + y = t ( y + 1)^(-2) - 1
@& We rearrange this with the substitution u = y + 1, so that u ' = y ', obtaining
u ' = -u + t u^2
then
u ' + u = t u^2.
This is a Bernoulli equation with n = 2.
Let v = u^m = u^(1 - n) = u^(-1).
The equation becomes
v ' - v = t.
Our integrating factor is e^(-t), giving us
(v e^(-t)) ' = t e^(-t).
Integrating both sides
v e^(-t) = -t e^(-t) - e^(-t) + c
so that
v = -t - 1 + c e^t,
u^(-1) = (-t - 1 + c e^t)
u = (-t - 1 + c e^t)^(-1) = 1 / (-t - 1 + c e^t)
and since u = y + 1
y + 1 = 1 / (-t - 1 + c e^t)
y = y + 1 = 1 / (-t - 1 + c e^t) - 1*@
@& You're on the right track with everything. Just a couple of minor glitches.
Check my notes and see if you agree.*@