General Solution

**Parton-Bill** · April 9th 2010, 05:46 AM

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

**Prove It** · April 9th 2010, 06:41 AM

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 $A$ .

**maddas** · April 9th 2010, 06:51 AM

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.

**Parton-Bill** · April 9th 2010, 08:06 AM

wow you guys truely rock lol

thanks very muchy guys

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