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Math Help - Having problem with this separation

  1. #1
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    Having problem with this separation

    I'm pretty sure I'm doing something wrong.
    the question is:
    Formulate the general solution to the differential equation. (express in y^2 in terms of x)

    dy/dx = e^x/y + 2y

    This is what I did doing so far.

    dy/dx = e^x/y + 2y

    dy/dx  - 2y = e^x/y
    dy/ydx  = e^x +2
    dy  = y(e^x +2)dx
    (1/y) dy  = (e^x +2)dx
    ln(y) = e^x + 2x
    y = e^(e^x+2x)<br />
    y^2 =( e^(e^x+2x))^2<br />

    the final answer just looks way too weird.
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  2. #2
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    Quote Originally Posted by konvos View Post
    I'm pretty sure I'm doing something wrong.
    the question is:
    Formulate the general solution to the differential equation. (express in y^2 in terms of x)

    dy/dx = e^x/y + 2y

    This is what I did doing so far.

    dy/dx = e^x/y + 2y

    dy/dx  - 2y = e^x/y
    dy/ydx  = e^x +2
    dy  = y(e^x +2)dx
    (1/y) dy  = (e^x +2)dx
    ln(y) = e^x + 2x
    y = e^(e^x+2x)<br />
    y^2 =( e^(e^x+2x))^2<br />

    the final answer just looks way too weird.
    First lets clean this up a bit

    If we multiply by y we get

    \displaystyle y\frac{dy}{dx}=e^{x}+2y^2

    Now make the substitution

    \displaystyle u=y^2 \implies \frac{du}{dx}=2y\frac{dy}{dx}

    Subbing this in gives

    \displaystyle \frac{1}{2}\frac{du}{dx}=e^{x}+2u \iff \frac{du}{dx}-4u=2e^{x}

    This equation can be solved via an integrating factor. Can you finish from here ?
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  3. #3
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    Just as a further explanation of TheEmptySet's post: the original DE is Bernoulli, and hence a substitution of the form u=y^{2} will render the equation linear in u.
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    First lets clean this up a bit

    If we multiply by y we get

    \displaystyle y\frac{dy}{dx}=e^{x}+2y^2

    Now make the substitution

    \displaystyle u=y^2 \implies \frac{du}{dx}=2y\frac{dy}{dx}

    Subbing this in gives

    \displaystyle \frac{1}{2}\frac{du}{dx}=e^{x}+2u \iff \frac{du}{dx}-4u=2e^{x}

    This equation can be solved via an integrating factor. Can you finish from here ?
    Thx for the help.
    The Y^2 is getting my confused. The formula I'm applying is


    P(x) = -4, hence  the derivative is -4x
    applying to both sides, I should get
    -4xy^2 = 2e^dx
    I just guessed that the y^2 would be squared, I tried looking for a example in my book and I couldn't find.
    so the final answer is :
    y^2 = -2e^x/4x but like I said... I just guessed on the y^2, so i'm not sure
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  5. #5
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    Quote Originally Posted by konvos View Post
    Thx for the help.
    The Y^2 is getting my confused. The formula I'm applying is
    Why are you applying that formula, when it doesn't apply? That formula is the form of a first-order linear DE, but that's not what you start out with. If you will employ TheEmptySet's substitution, you will get a first-order linear DE in u, at which point you can employ the integrating factor method.
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  6. #6
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    Quote Originally Posted by Ackbeet View Post
    Why are you applying that formula, when it doesn't apply? That formula is the form of a first-order linear DE, but that's not what you start out with. If you will employ TheEmptySet's substitution, you will get a first-order linear DE in u, at which point you can employ the integrating factor method.
    I'm confused by the substitution but if I substitute I'll get
    2y(dy/dx) -4y^2= 2e^x
    2y(dy/dx - 2y) = 2e^x
    I'm not sure if the formula could be applied there.
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  7. #7
    Behold, the power of SARDINES!
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    Quote Originally Posted by konvos View Post
    I'm confused by the substitution but if I substitute I'll get
    2y(dy/dx) -4y^2= 2e^x
    2y(dy/dx - 2y) = 2e^x
    I'm not sure if the formula could be applied there.
    You have not done the substition yet! If you look back in post #2 I did the substition. The NEW ODE has u as the dependant variable and x as the independant variable.

    Use the formula on this equation.
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