Question

how do you solve dy/dx=(y−1)(y+1) if the solution passes through the point (x,y)=(4,0)(x,y)=(4,0)?

I got y=(1+e^(2x-8))/(1-e^(2x-8)) but it was wrong:(

Answer 1

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