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Homework Help: ODE Problem, shifting origin

  1. Sep 10, 2016 #1


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    1. The problem statement, all variables and given/known data
    Find a particular solution to:
    (3x+2y+3)dx - (x+2y-1)dy = 0, y(-2) = 1

    The answer to this problem as presented in the book ODE by Tenenbaum is the following:

    2. Relevant equations
    I will be shifting the origin to try to compute this problem.

    3. The attempt at a solution


    \frac {dy} {dx} = -\frac {(3x+2y+3)} {(-x-2y+1)}[/tex]

    The system of lines is as follows:

    [tex] y = -\frac {3x} {2}-\frac {3} {2}[/tex]
    [tex] y = -\frac {x} {2}+\frac {1} {2}[/tex]

    This system has a solution of x = -2 and y = 3/2

    To shift the origin I set:

    [tex] x = \overline{x}-2, y = \overline{y}+\frac{3} {2} [/tex]
    [tex] \overline{x} = x+2, \overline{y} = y-\frac{3} {2} [/tex]

    Now to start solving the problem we input the shifted values of x and y.

    [tex] (3(\overline{x}-2)+2(\overline{y}+\frac{3} {2}) + 3)d\overline{x} + (-(\overline{x}-2)-2(\overline{y}+\frac{3} {2})+1)d\overline{y} = 0 [/tex]

    This should lead (I've checked my calculations here multiple times) to:

    [tex] (3\overline{x}+2\overline{y})d\overline{x}+(-\overline{x}-2\overline{y})d\overline{y}=0[/tex]

    Now I'm ready to solve this using methods that I had previously learned.

    [tex]Set y=u\overline{x}, d\overline{y} = ud\overline{x}+\overline{x}du[/tex]

    [tex] (3\overline{x}+2u\overline{x})d\overline{x}+(-\overline{x}-2(u\overline{x}))(ud\overline{x}+\overline{x}du)=0[/tex]

    Calculating this and simplifying we get:

    [tex] (-2u^2\overline{x}+3\overline{x}+u\overline{x})d\overline{x}+(-\overline{x}^2-2u\overline{x}^2)du = 0[/tex]

    Now, with this, I divide everything by x-bar^2. I then put it into seperable form by dividing through by (-2u^2+3+u). This becomes:

    [tex] \frac {d\overline{x}} {\overline{x}} + \frac {(-1-2u)} {(-2u^2+3+u)} du = 0 [/tex]

    A separable equation! Now I can integrate. I integrate and net :

    [tex] log(\overline{x}) + \frac {1} {5}(4log(3-2u)+log(u+1)) + c [/tex]

    I try to manipulate the solution to appear closer to the given solution as presented in the book. So I multiply by e throughout the entire solution. I then raise everything to the 5th power.

    [tex] \overline{x}^5 + (3-2u)^4(u+1) = 0 [/tex]
    We recall that [tex] u=\frac {\overline{y}} {\overline{x}} [/tex]

    plugging in this, and then also plugging in the values for x-bar and y-bar i net:

    [tex] (x+2)^5+(3-\frac{2y-3} {x+2})^4(\frac{y-\frac{3} {2}} {x+2} +1) = C [/tex]

    Well, this is bad. Using the initial condition y(-2)=1 if we let x = -2 we get a fraction with 0 in the denominator.

    What have I done wrong? :/
  2. jcsd
  3. Sep 10, 2016 #2

    Ray Vickson

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    If ##\hat{y} = u \hat{x}## does not work, try instead ##\hat{x} = u \hat{y}##.

    Basically, a point on the ##\hat{y}##-axis cannot be described as lying on a line ##\hat{y} = u \hat{x}## with finite ##u##, but it does lie on a line ##\hat{x} = u \hat{y}## with finite ##u##.
    Last edited: Sep 10, 2016
  4. Sep 10, 2016 #3


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    Hi Ray, I appreciate the comment!

    I'm at work right now, but I worked it out the way you suggested:

    [tex] \overline{x} = u\overline{y}[/tex] and [tex] d\overline{x} = ud\overline{y}+\overline{y}du[/tex]

    After plugging it all into the equation and simplifying I get a seperable expression of the following:

    [tex] \frac{1} {y}d\overline{y} + \frac{3u+2} {3u^2+u-2}du = 0[/tex]

    While this looks good, plugging in the integral into symbolab gives me a very, very long expression. here

    This leads me to believe that this way is not working properly either. Admittedly, I did not bother plugging back in everything as that would be quite the task with that integral solution.
  5. Sep 10, 2016 #4

    Ray Vickson

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    I clicked on the link you provided to see what symbolab gives, and I am not convinced it is correct. Maple gets a pretty simple result:
    $$\int \frac{3u+2}{3u^2+u-2} \, du = \frac{1}{5} \ln(u+1) +\frac{4}{5} \ln(3u-2) $$
    You can check that this is correct by taking the derivative. Can you check the symbolab result like that? Also: Instead of using symbolab, why not try Wolfram Alpha?

    BTW: the integration can be done easily manually, because the denominator factors as ##(u+1)(3u-2)##, so the integrand can be written in partial fractions.
    Last edited: Sep 10, 2016
  6. Sep 11, 2016 #5


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    You are absolutely right.

    I did the integration manually using partial fraction expansion and that made it a reasonable integral.

    I was able to simplify it to:

    [tex](y-\frac{3} {2})^5+(\frac{x+2} {y-\frac{3} {2}}) + (\frac{3x+6} {y-\frac{3} {2}} -2)^4 = C [/tex]

    So, it's an answer. I feel like I made no mistakes throughout the calculation. I am 'happy' with this answer. But I'm not sure if i can manipulate it to look like the given answer from the book.
    Last edited: Sep 11, 2016
  7. Sep 11, 2016 #6

    Ray Vickson

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    You seem to have written that ##\ln(A) + \ln(B) + \ln(C) = k## implies ##A + B + C = k'##, but that is false: it should be ##A B C = k'## (product, not sum).
  8. Sep 11, 2016 #7


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    What a mistake and what a way to wake up to the morning! Problem Solved. My mistakes... noted.

    Kind regards.
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