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Thread: Solving Differential Equation

  1. #1
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    Solving Differential Equation

    Hi,

    I am struggling to finish this question and would love some guidance please:

    Question: Solve for the following differential equation for the original equation:$\displaystyle y' = \frac{-2x-y^2}{2xy-5}$

    $\displaystyle y' = \frac{-2x-y^2}{2xy-5}$

    $\displaystyle \frac{dy}{dx}= \frac{-2x-y^2}{2xy-5}$

    $\displaystyle (2xy-5)dy = (-2x-y^2)dx$

    $\displaystyle 2xydy - 5dy = -2xdx -y^2dx$ (assuming that I'm correct so far, what do I do now?)
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by sparky View Post
    Question: Solve for the following differential equation for the original equation:$\displaystyle y' = \frac{-2x-y^2}{2xy-5}$
    Are you asked to solve this or put it into a more standard form?

    -Dan
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    Quote Originally Posted by topsquark View Post
    Are you asked to solve this or put it into a more standard form?

    -Dan
    Thanks for the reply. I'm asked to solve it "for the original equation". This sounds a little confusing to me.
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    Quote Originally Posted by topsquark View Post
    Are you asked to solve this or put it into a more standard form?

    -Dan
    Thanks for the reply. I'm asked to solve it "for the original equation". This sounds a little confusing to me.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by sparky View Post
    Hi,

    I am struggling to finish this question and would love some guidance please:

    Question: Solve for the following differential equation for the original equation:$\displaystyle y' = \frac{-2x-y^2}{2xy-5}$

    $\displaystyle y' = \frac{-2x-y^2}{2xy-5}$

    $\displaystyle \frac{dy}{dx}= \frac{-2x-y^2}{2xy-5}$

    $\displaystyle (2xy-5)dy = (-2x-y^2)dx$

    $\displaystyle 2xydy - 5dy = -2xdx -y^2dx$ (assuming that I'm correct so far, what do I do now?)
    Okay, I saw a similar example in one of your other posts. You are good for as far as you have gotten. Regroup:
    $\displaystyle (2xydy + y^2dx) - 5dy + 2xdx = 0$

    The $\displaystyle 2xy dy + y^2 dx$ is a perfect differential. What is it?

    -Dan
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    Quote Originally Posted by topsquark View Post
    You are good for as far as you have gotten.
    Ok this is very encouraging.

    The $\displaystyle 2xy dy + y^2 dx$ is a perfect differential. What is it?
    I am still trying to figure out why you grouped together $\displaystyle (2xydy + y^2dx)$ rather than $\displaystyle (2xydy-5dy)$. Is it because we of differentiating with respect to y (or is it x, I'm a little confused I must admit)?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by sparky View Post
    Ok this is very encouraging.



    I am still trying to figure out why you grouped together $\displaystyle (2xydy + y^2dx)$ rather than $\displaystyle (2xydy-5dy)$. Is it because we of differentiating with respect to y (or is it x, I'm a little confused I must admit)?
    Acutally I'm going to integrate over something different. Note that
    $\displaystyle (2xydy + y^2dx) = d(xy^2)$

    So when it is time to take integrals I will do it like this:
    $\displaystyle \displaystyle \int d(xy^2) = xy^2 + C$

    The reason why I grouped it the way I did was so I could get rid of the y dx and x dy terms.

    -Dan
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