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First order separable differential equation

  1. Sep 12, 2012 #1
    1. The problem statement, all variables and given/known data

    x[itex]\frac{dy}{dx}[/itex] = 4y


    2. Relevant equations

    I'm not sure if there is a specific equation for these type of problems. My professor just says to separate the two different variables and then integrate them with respect to x.


    3. The attempt at a solution

    [itex]\frac{1}{4y}[/itex] [itex]\frac{dy}{dx}[/itex] = [itex]\frac{1}{x}[/itex]

    ∫[itex]\frac{1}{4y}[/itex] [itex]\frac{dy}{dx}[/itex] dx =∫[itex]\frac{1}{x}[/itex] dx

    left side:

    ∫[itex]\frac{1}{4y}[/itex] [itex]\frac{dy}{dx}[/itex] dx

    u = y
    du = [itex]\frac{dy}{dx}[/itex] dx

    [itex]\frac{1}{4}[/itex]∫[itex]\frac{1}{u}[/itex]du

    ln(4y) + [itex]C_{y}[/itex]


    right side:

    ∫[itex]\frac{1}{x}[/itex] dx

    ln(x) + [itex]C_{x}[/itex]

    ln(4y) = ln(x) + C

    Do i have to continue or could I just stop here?

    4y = [itex]e^{x}[/itex] * [itex]e^{c}[/itex]

    y = [itex]\frac{e}{4}^{x}[/itex] * [itex]\frac{e}{4}^{c}[/itex]
     
    Last edited by a moderator: Sep 12, 2012
  2. jcsd
  3. Sep 12, 2012 #2

    Mark44

    Staff: Mentor

    Please don't use SIZE tags.
    This is never a good substitution, because all you're doing is switching to a different letter.

    You should have ended up with (1/4)ln|y| on the left side.

    You made the problem considerably harder than it is by not immediately moving the constant outside the integral, and by dragging dy/dx along.

    ∫1/(4y)dy = (1/4)∫dy/y
    You have another mistake above. When you exponentiate the right side, you get eln(x) + C = eln(x) * eC. That first factor can be greatly simplfied.
     
  4. Sep 12, 2012 #3
    Sorry, I didn't know that wasn't allowed.

    I was kind of following the example my teacher did that was similar to this problem.

    I remember my calc 2 teacher telling me the integral of 1/Ax is ln(Ax). So that's why I thought it was ln(4y). When I did u substitution I thought it was kind of strange how the 4 ends up back inside ln. I guess I heard him wrong


    Yeah I know that e^ln(x) is just x i just copied it down incorrectly when i was posting it here.


    So would it be,

    (1/4)ln(y) = ln(x) +C

    ln(y) = 4ln(x) +C

    y = 4x +C

    I don't think I exponentiated this correctly. If it were e^ln(4x) then it would be 4x.

    Can e^4(lnx) be written as e^4 * e^ln(x)?

    I don't think that's right either because e^(ln(x) +c) is e^ln(x) * e^c. I'm confused.
     
    Last edited: Sep 12, 2012
  5. Sep 13, 2012 #4

    Mark44

    Staff: Mentor

    It's not that it isn't allowed - it's just not needed most of the time.
    Either your teacher told you something that's wrong or you misheard. You can check this formula by differentiating. d/dx(ln(Ax)) = 1/(Ax) * A = 1/x ≠ 1/(Ax)
    Above, it would be 4C, but that's just a different constant. Some people just write C' to indicate that it's different from C. For this problem C' = 4C.
    No, that's incorrect.

    Starting from the equation before this one...

    ln(y) = 4ln(x) + C'

    eln(y) = e4ln(x) + C' = e4ln(x) * eC'
    => y = eC' * e4ln(x) = Aeln(x4) = Ax4

    Note that eC' is just another constant, so call it A.

    If you really want to keep track, A = eC' = e4C

    Strictly speaking, there should be absolute values involved on both sides. The real formulas are
    $$ \int \frac{dy}{y} = ln|y| + C$$
    and
    $$ \int \frac{dx}{x} = ln|x| + C$$
    I'm not sure what you mean by e^4(lnx). Is this e^4 * ln(x) or e^(4ln(x))?
     
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