I need help in solving this ivp differential equation
$$\frac{dx}{dt} = k(a-x)^2 (b-\frac{x}{2})$$
$a=0.0010$
$b =0.0041$
$k=7.13x10^3$
with ivp $x(0) = 0$
by python solution, the answer should be
by manually solve the equation before substitute it with ivp, i get this
Am I on the right track, because if I replace with the ivp value, $x(0)=0$ to get C value, I get different answer as python
I'm stuck in solving this, any help will be appreciated!
As
$$\frac{dx}{dt} = k(a-x)^2 \left(b-\frac{x}{2}\right)$$
is separable we have
$$ -\frac{2 ((a-x) \log (x-2 b)+(x-a) \log (x-a)+a-2 b)}{k (a-2 b)^2 (a-x)}=t+c_0 $$
or
$$ \frac{1}{2} k (a-2 b)^2 (t+c_0)+\frac{a-2 b}{a-x}-\log (x-a)+\log (x-2 b)=0 $$
or
$$ e^{\frac{a-2 b}{a-x}}\frac{x-2b}{x-a}-e^{-\frac{1}{2} k (a-2 b)^2 (t+c_0)}=0 $$
now making $(t=0, x=0)$ we have
$$ e^{\frac{a-2 b}{a}}\frac{2b}{a}-e^{-\frac{1}{2} k (a-2 b)^2 c_0}=0 $$
so
$$ c_0 = \frac{2 \log \left(\frac{a e^{\frac{2 b}{a}-1}}{2 b}\right)}{k (a-2 b)^2} $$