Respuesta :
The solution to the differential equation is:
[tex]y = \frac{1 - e^{2x - 8}}{1 + e^{2x - 8}}[/tex]
What is the differential equation?
It is given by:
[tex]\frac{dy}{dx} = (y - 1)(y + 1)[/tex]
Applying separation of variables, we have that:
[tex]\frac{dy}{(y - 1)(y + 1)} = dx[/tex]
[tex]\int \frac{dy}{(y - 1)(y + 1)} = \int dx[/tex]
For the first integral, we use partial fractions, that is:
[tex]\int \frac{dy}{(y - 1)(y + 1)} = \int \frac{A}{y - 1} dy + \int \frac{B}{y + 1} dy[/tex]
In which:
[tex]A = \frac{1}{y+1}_{y = 1} = \frac{1}{2}[/tex]
[tex]B = \frac{1}{y - 1}_{y = -1} = -\frac{1}{2}[/tex]
Then:
[tex]\int \frac{dy}{(y - 1)(y + 1)} = \int \frac{1}{2(y - 1)} dy - \int \frac{1}{2(y + 1)} dy[/tex]
[tex]\int \frac{dy}{(y - 1)(y + 1)} = \frac{1}{2}\ln{(y - 1)} - \frac{1}{2}\ln{(y + 1)} = \frac{1}{2}(\ln{(y - 1)} - \ln{(y + 1)})[/tex]
Then, coming back to the separation of variables:
[tex]\frac{1}{2}(\ln{(y - 1)} - \ln{(y + 1)}) = x + K[/tex]
In which K is the constant of integration.
Then:
[tex]2x + K = \ln{(y - 1)} - \ln{(y + 1)}[/tex]
[tex]2x + K = \ln{\frac{y - 1}{y + 1}}[/tex]
[tex]e^{2x + K} = e^{\ln{\frac{y - 1}{y + 1}}}[/tex]
[tex]Ke^{2x} = \frac{y - 1}{y + 1}[/tex]
Since it passes through point (4,0), when [tex]x = 4, y = 0[/tex], and this is used to find K.
[tex]Ke^{8} = \frac{0 - 1}{0 + 1}[/tex]
[tex]K = -e^{-8}[/tex]
Then:
[tex]Ke^{2x} = \frac{y - 1}{y + 1}[/tex]
[tex]-e^{-8}e^{2x} = \frac{y - 1}{y + 1}[/tex]
[tex]-e^{2x - 8} = \frac{y - 1}{y + 1}[/tex]
[tex]y - 1 = -ye^{2x - 8} - e^{2x - 8}[/tex]
[tex]y + ye^{2x - 8} = 1 - e^{2x - 8}[/tex]
[tex]y(1 + e^{2x - 8}) = 1 - e^{2x - 8}[/tex]
[tex]y = \frac{1 - e^{2x - 8}}{1 + e^{2x - 8}}[/tex]
You can learn more about differential equations at https://brainly.com/question/14318343
