Assignment 9

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course Mth 279

6/20 9:30

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(1-y)

y’ = 2ty - 2ty^2

y’ - 2ty = -2ty^2

p(t) = -2t q(t) = -2t n = 2

v’ + (1 -n)p(t)v = (1-n)q(t)

v’ + (1 -2)(-2t)v = (1-2)(-2t)

v’ + 2tv = 2t

p(t) = 2t int(p(t)) = 2t^2/2 = t^2

e^t^2

e^(t^2)v’ + 2te^(t^2)v = 2te^(t^2)

integrate with respect to t on both sides

e^(t^2)v = e^t^2 + C

v = 1 + C/(e^(t^2))

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

y = v^(1/(1-n)) = v^-1

y = (1+Ce^(-t^2))^-1

-1 = 1/(1+Ce^(-0^2))

-(1+C) = 1

C = -2

y = 1/(1+2e^(-t^2))

confidence rating #$&*:

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1

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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:

p(t) = -1 q(t) = t n=(1/3)

v’ -(2/3)v = (2/3)t

p(t) = -2/3 int(p(t)) = -(2/3)t

e^((-2/3)t)

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

integrate by parts and then u-sub on right side of equation

u = t du = dt dv=e^((-2/3)t) v=-(3/2)e^((-2/3)t)

(2/3)((-3/2)te^((-2/3)t) - int((-3/2)e^((-2/3)t) let w = (-2/3)t dw = -2/3

e^((-2/3)t)v = -te^((-2/3)t) - (3/2)e^((-2/3)t) + C

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

y = v^(3/2)

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

-9 = (0 - (3/2) + Ce^0)^3/2

-9 = -(3/2) + C)^(3/2)

-9^(2/3) = -(3/2) +C

C = (-9)^(2/3) + (3/2) = 5.83

y= (-t -(3/2) + 5.83e^((2/3)t))^(3/2)

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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 ' + 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)

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confidence rating #$&*:

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1

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Given Solution:

Unsure about math when calculating C value. Couldn’t find my mistake if there is one

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Self-critique (if necessary):

Unsure about math when calculating C value. Couldn’t find my mistake if there is one

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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:

Let z = y+1

therefore z’ = y’

z’ = -z + tz^-2

z’ + z = tz^-2

p(t) = 1 q(t)= t n=-2

v’ +3v = 3t

p(t) =3 int(p(t)) = 3t

e^(3t)v’ + 3e^(3t)v = 3e^(3t)t

integrate using by parts on right side of equation

e(3t)v =te^(3t) - (1/3)e^(3t) + C

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

z = v^(1/3)

z = (t - (1/3) + Ce^(-3t))^(1/3)

subbing back in z value

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

y = (t - (1/3) + Ce^(-3t))^(1/3) -2

confidence rating #$&*:

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2

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Given Solution:

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Self-critique (if necessary):

Pretty easy once I saw from the book the hint to let z=y+1. I don’t think I would have thought of that.

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Self-critique rating:"

&#This looks good. See my notes. Let me know if you have any questions. &#