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Math Help - Solving for Original Equation

  1. #1
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    Solving for Original Equation

    Is my working and answer correct for the following question?

    Question: Solve for the following differential equation for the original equation:

    y' = \frac{2x-3y}{3x+8y}

    Here is my working:

    \frac{dy}{dx} = \frac{2x-3y}{3x+8y}

    (3x+8y)dy = (2x-3y)dx

    \int 3xdy + \int 8ydy = \int 2xdx - \int 3ydx (why does the intergral of 3ydx = 0?)

    \frac{3x^2}{2} + \frac{8y^2}{2} = \frac{2x^2}{2} - 0 + C

    3x^2 + 8y^2 - 2x^2 - C = 0

    x^2 + 8y^2 - C = 0
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  2. #2
    A Plied Mathematician
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    Two comments:

    1. This method doesn't work, since you don't know what y is a priori. If you did, the DE would be solved!
    2. The DE is homogeneous. Use the appropriate technique.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by sparky View Post
    Is my working and answer correct for the following question?

    Question: Solve for the following differential equation for the original equation:

    y' = \frac{2x-3y}{3x+8y}

    Here is my working:

    \frac{dy}{dx} = \frac{2x-3y}{3x+8y}

    (3x+8y)dy = (2x-3y)dx

    \int 3xdy + \int 8ydy = \int 2xdx - \int 3ydx (why does the intergral of 3ydx = 0?)

    \frac{3x^2}{2} + \frac{8y^2}{2} = \frac{2x^2}{2} - 0 + C

    3x^2 + 8y^2 - 2x^2 - C = 0

    x^2 + 8y^2 - C = 0
    Just like in your other problems, it's a matter of grouping.
    (3x + 8y) dy = (2x - 3y) dx

    3x dy + 8y dy = 2x dx - 3y dx

    Regroup:
    (3x dy + 3y dx) + 8y dy = 2x dx

    Factor the 3:
    3 (x dy + y dx) + 8y dy = 2x dx

    Now I showed you how to handle x dy + y dx in a different thread. Do you remember what I did?

    -Dan
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