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Math Help - Separable Equation Problem.

  1. #1
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    Separable Equation Problem.

    Having not studied calculus for a couple of years I am a tad rusty. Was hoping someone could give me a kick start.

    The question I need to solve is:

    dy/dx = [(4y + 5)/(2x + 3)]^2

    Any help would be much appreciated...


    Regards.
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  2. #2
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    Re: Separable Equation Problem.

    Quote Originally Posted by paulbk108 View Post
    Having not studied calculus for a couple of years I am a tad rusty. Was hoping someone could give me a kick start.

    The question I need to solve is:

    dy/dx = [(4y + 5)/(2x + 3)]^2

    Any help would be much appreciated...


    Regards.
    \frac{dy}{dx}=\frac{(4y+5)^2}{(2x+3)^2} \Rightarrow \int \frac{1}{(4y+5)^2} \, dy=\int \frac{1}{(2x+3)^2} \, dx

    use substitutions  u=4y+5 and v=2x+3 to solve integrals .
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  3. #3
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    Re: Separable Equation Problem.

    So is that then integration by parts?


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  4. #4
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    Re: Separable Equation Problem.

    Quote Originally Posted by paulbk108 View Post
    So is that then integration by parts?


    Regards.
    No , if you make substitution you should get integral of the form :

    \int u^{-2}\, du
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  5. #5
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    Re: Separable Equation Problem.

    So;

    int [u^-2} = int [v^-2]

    = (u^-1)/-1 = (v^-1)/-1

    = [(4y + 5)^-1]/-1 = [(2x + 3)^-1]/-1 + c

    Doesn't seem correct...?


    Regards.
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  6. #6
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    Re: Separable Equation Problem.

    Quote Originally Posted by paulbk108 View Post
    So;

    int [u^-2} = int [v^-2]

    = (u^-1)/-1 = (v^-1)/-1

    = [(4y + 5)^-1]/-1 = [(2x + 3)^-1]/-1 + c

    Doesn't seem correct...?


    Regards.
    Note that :  du=4dy and dv=2dx
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  7. #7
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    Re: Separable Equation Problem.

    u = 4y + 5
    => du/dy = 4
    =>dy = du/4

    => int [dy/(4y + 5^2)] = 1/4 int [du/u^2] = ...?

    v = 2x + 3
    =>dv/dx = 2
    =>dx = dv/2

    => int [dx/(2x + 3)^2] = 1/2 int [dv/v^2] = ...?

    Is that correct?


    Regards.
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  8. #8
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    Re: Separable Equation Problem.

    yes , it is..
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