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Math Help - General Solution

  1. #1
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    General Solution

    I must have missed this lecture, as with the particular solution, I am stumped, I have information but without the teaching its just garble.

    Please advise on the following General Solution:

    dy/dx -4y^2 = 2y

    Many thanks
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  2. #2
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    This DE is separable.

    \frac{dy}{dx} - 4y^2 = 2y

    \frac{dy}{dx} = 4y^2 + 2y

    \frac{dy}{dx} = 2y(2y + 1)

    \frac{1}{2y(2y + 1)}\,\frac{dy}{dx} = 1

    \int{\frac{1}{2y(2y + 1)}\,\frac{dy}{dx}\,dx} = \int{1\,dx}

    \int{\frac{1}{2y(2y + 1)}\,dy} = \int{1\,dx}.


    To evaluate the LHS, you need to use partial fractions.

    \frac{1}{2y(2y + 1)} = \frac{A}{2y} + \frac{B}{2y + 1}

     = \frac{A(2y + 1) + 2By}{2y(2y + 1)}.


    Therefore A(2y + 1) + 2By = 1

    2Ay + A + 2By = 1

    (2A + 2B)y + A = 0y + 1.


    Therefore A = 1 and 2A + 2B = 0

    2 + 2B = 0

    B = -1.


    So \frac{1}{2y(2y + 1)} = \frac{1}{2y} - \frac{1}{2y + 1}.


    So the DE becomes

    \int{\frac{1}{2y} - \frac{1}{2y + 1}\,dy} = \int{1\,dx}

    \frac{1}{2}\ln{|y|} - \frac{1}{2}\ln{|2y + 1|} + C_1 = x + C_2

    \frac{1}{2}\ln{\left|\frac{y}{2y + 1}\right|} = x + C_2 - C_1

    \ln{\left|\frac{y}{2y + 1}\right|} =  2x + C where C = 2C_2 - 2C_1

    \ln\left|\frac{1}{2} - \frac{1}{2(2y + 1)}\right| = 2x + C

    \left|\frac{1}{2} - \frac{1}{2(2y + 1)}\right| = e^{2x + C}

    \left|\frac{1}{2} - \frac{1}{2(2y + 1)}\right| = e^Ce^{2x}

    \frac{1}{2} - \frac{1}{2(2y + 1)} = Ae^{2x} where A = \pm e^C

    \frac{1}{2(2y + 1)} = \frac{1}{2} - Ae^{2x}

    \frac{1}{2(2y + 1)} = \frac{1 - 2Ae^{2x}}{2}

    2(2y + 1) = \frac{2}{1 - 2Ae^{2x}}

    2y + 1 = \frac{1}{1 - 2Ae^{2x}}

    2y = \frac{1}{1 - 2Ae^{2x}} - 1

    2y = \frac{1 - (1 - 2Ae^{2x})}{1 - 2Ae^{2x}}

    2y = \frac{2Ae^{2x}}{1 - 2Ae^{2x}}

    y = \frac{Ae^{2x}}{1 - 2Ae^{2x}}.


    Now if you have been given a boundary condition, you can solve for A.
    Last edited by Prove It; April 9th 2010 at 06:55 AM.
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  3. #3
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    Rearranging, {2y-\dot{y} \over y^2} = -4. Multiply by h, which is an integrating factor of sorts [this just gives us something to work with]. We have {2hy-h\dot{y} \over y^2} = -4h.

    Notice that the LHS looks like the result of the quotient rule, so look for f and g with \frac{\mathrm{d}}{\mathrm{d}x} ({f \over g}) = {2hy-h\dot{y} \over y^2}. Differentiating \frac{\mathrm{d}}{\mathrm{d}x} ({f \over g}) = {\dot{f}g - f\dot{g} \over g^2} = {2hy-h\dot{y} \over y^2}.

    Looking at the denominator, set g := y. Then compare the numerators and see that f = h and \dot{f} = 2h. So we have:

    \frac{\mathrm{d}}{\mathrm{d}x} ({h \over y}) = -4h where h satisfies \dot{h}=2h

    h is easy: h(x) = C e^{2x}. For the other, integrate from a to b to get

    \int_a^b\mathrm{d} ({h \over y}) = {h\over y}\Big|_a^b = \int_a^b-4h\,\mathrm{d}x = -4C \int_a^b e^{2x} \,\mathrm{d}x = -2C (e^{2b}-e^{2a})

    Ce^{2b}/y(b) = -2C (e^{2b}-e^{2a}) + Ce^{2a}/y(a)

    y(b) = \frac{e^{2b}}{-2 (e^{2b}-e^{2a}) + e^{2a}/y(a)}

    which will simplify if you like.
    Last edited by maddas; April 9th 2010 at 06:54 AM. Reason: sign error lol
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  4. #4
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    wow you guys truely rock lol thanks very muchy guys
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