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Math Help - general solution to DE

  1. #1
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    general solution to DE

    dy/dx +2y/x = 4x

    find general solution

    dy/dx= 4x + x/2y

    2y dy = 4x + x dx

    2y dy = 5x dx

    integrate

    y^2 = 2.5 x^2 + C
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  2. #2
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    Quote Originally Posted by harveyo View Post
    dy/dx +2y/x = 4x

    find general solution

    dy/dx= 4x + x/2y

    2y dy = 4x + x dx

    2y dy = 5x dx

    integrate

    y^2 = 2.5 x^2 + C
    What you have done is not correct.

    Your DE is not separable, you need to use the integrating factor method.

    Do you know this method?

    Read this http://www.mathhelpforum.com/math-he...equations.html
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  3. #3
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    y' + 1/x 2y = 4x

    so I(x) = e intergral 1/x dx = e ln x = x

    now

    x y' + 2y = 4x^2
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  4. #4
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    Quote Originally Posted by harveyo View Post
    y' + 1/x 2y = 4x

    so I(x) = e intergral 1/x dx = e ln x = x

    now

    x y' + 2y = 4x^2
    make it y' + \frac{2}{x} ~y = 4x

    now find I(x) = e^{\int \frac{2}{x}~dx}
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  5. #5
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    = e ln x^2 = x^2 ???

    now

    x^2 dy/dx + 2/x ^2 y = 4x^3

    (x^2 y )' = 4x^3

    y = x^2
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  6. #6
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    Quote Originally Posted by harveyo View Post
    = e ln x^2 = x^2 ???

    now
    Yep, great work, now multipling x^2 through the whole equation you get

    x^2 \times (y' + \frac{2}{x} ~y = 4x) = y'x^2+2xy= 4x^3

    and

     y'x^2+2xy= 4x^3

    by the product rule

     x^2y= \int 4x^3~dx

    Now can you take it from here?
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