General solution for ODE $\frac{dy}{dt} = (1-y)(3-y)(5-t)$ Can someone please explain me highlighted steps?

Question

$\frac{dy}{dt} = (1-y)(3-y)(5-t)$

ODE is seperable with equilibrium $y=1,3$ for $y\neq1, y\neq3$ we have

$\frac{1}{(y-1)(y-3)}\frac{dy}{dt} = (5-t)$

$\frac{1}{2}\frac{1}{(y-1)(y-3)}\frac{dy}{dt} = (5-t)$ $\color{red}{*}$

$\int\frac{1}{2}\left(\frac{1}{y-3}-\frac{1}{y-1}\right) dy = \int(5-t)dt$ $\color{blue}{*}$

$\frac{1}{2}ln|{y-3}|-\frac{1}{2}ln|{y-1}|= 5t-\frac{1}{2}t^2+\frac{1}{2}C$

$ln|\frac{{y-3}}{{y-1}}|=10t-t^2+C$

$\frac{{y-3}}{{y-1}}=Kexp(10-t^2)$

$y=\frac{Kexp(10-t^2)-3}{Kexp(10-t^2)-1}$

$\color{red}{*}$"I did not understand this step why they put $1/2$ here"

$\color{blue}{*}$"the two terms in bracket dot product should be addition not substraction "

Answer 1

Its by mistake. Step with red mark doesn't need to contain $\frac 12$

And blue mark step -

We have,

$\frac{1}{(y-3)} - \frac{1}{(y-1)}$

On simplifying,

$= \frac{(y-1)-(y-3)}{(y-3)(y-1)}$

$= \frac{y-1-y+3}{(y-3)(y-1)}$

$= \frac{2}{(y-3)(y-1)}$

If you multiply it by $\frac 12$

We have,

$= \frac{1}{(y-3)(y-1)}$

Way to do this -

Following is Equation (1)

$\frac{1}{(y-3)(y-1)} = \frac{A}{(y-3)} + \frac{B}{(y-1)}$

$1 = A(y-1) + B(y-3)$

From expression $y-1=0$ then $y=1$, we have

$1 = A(1-1) + B(1-3)$

$1 = B(-2)$

$B = \frac{1}{-2}$

From expression $y-3=0$ then $y=3$, we have

$1 = A(3-1) + B(3-3)$

$1 = A(2)$

$A = \frac{1}{2}$

Putting value of A and B in equation (1),

$\frac{1}{(y-3)(y-1)} = \frac{\frac 12}{(y-3)} + \frac{\frac 1{-2}}{(y-1)}$

$\frac{1}{(y-3)(y-1)} = \frac 12 \left( \frac{1}{(y-3)} - \frac{1}{(y-1)}\right)$