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Thread: 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.

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

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

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

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

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

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


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

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

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


    Therefore $\displaystyle A(2y + 1) + 2By = 1$

    $\displaystyle 2Ay + A + 2By = 1$

    $\displaystyle (2A + 2B)y + A = 0y + 1$.


    Therefore $\displaystyle A = 1$ and $\displaystyle 2A + 2B = 0$

    $\displaystyle 2 + 2B = 0$

    $\displaystyle B = -1$.


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


    So the DE becomes

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

    which will simplify if you like.
    Last edited by maddas; Apr 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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