# Thread: solving differential equation

1. ## solving differential equation

1. Find the general solution of the following differential equations:
(a) dy/dx = y(1-y)
(b) dy/dx = 3 -2y

what i tried to do was move them to LHS....

[ $\frac{1}{y(1-y)}$]( $\frac{dy}{dx}$)=1

Integrate wrt x,

$\int$ $\frac{1}{y(1-y)}$ dy= $\int$1 dx

then...i'm stuck...

as for (b)
2y + dy/dx =3
....I couldn't figure it out how to express y in terms of x eventually.

any help is much appreciated.

2. The RHS is easy, $\displaystyle \int{1\,dx} = x + C_1$.

The LHS is solved using Partial Fractions:

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

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

$\displaystyle A(1-y)+By=1$

$\displaystyle A-Ay+By =1$

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

$\displaystyle B-A=0$ and $\displaystyle A=1$.

So $\displaystyle A=1, B=1$.

Therefore the LHS:

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

$\displaystyle = \ln{|y|} + \ln{|1-y|} + C_2$

$\displaystyle =\ln{\left|y(1-y)\right|} + C_2$.

So your DE solution is

$\displaystyle \ln{|y(1-y)|} +C_2 = x+C_1$

$\displaystyle \ln{|y(1-y)|} = x + C$ where $\displaystyle C = C_1-C_2$

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

$\displaystyle |y(1-y)| = e^Ce^x$

$\displaystyle y(1-y) = Ae^x$ where $\displaystyle A = \pm e^C$

$\displaystyle y - y^2 = Ae^x$

$\displaystyle y^2 - y = -Ae^x$

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

$\displaystyle \left(y-\frac{1}{2}\right)^2 = \frac{1}{4} - Ae^x$

$\displaystyle y-\frac{1}{2} = \pm \sqrt{\frac{1}{4} - Ae^x}$

$\displaystyle y - \frac{1}{2} = \pm \sqrt{\frac{1-4Ae^x}{4}}$

$\displaystyle y - \frac{1}{2} = \pm \frac{\sqrt{1-4Ae^x}}{2}$

$\displaystyle y = \frac{1 \pm \sqrt{1 - 4Ae^x}}{2}$.

3. Setting $u= y-\frac{1}{2}$ You obtain...

$\displaystyle \int \frac{dy}{y\ (1-y)} = \int \frac{du}{\frac{1}{4} - u^{2}} = 2\ \tanh^{-1} 2 u + c$

Merry Christmas from Italy

$\chi$ $\sigma$