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