$100\frac{dp}{dt}=p(100-p)$, find $p(t)$ So far what I have done:


*

*Recognised this is a separable ODE

*Rearranged into the form


$$100\int\frac1{p(100-p)}\ dp=\int1\ dt$$
However I am not sure how to proceed from here, I am having trouble integrating the LHS of the equation. Whatever I get seems too messy to isolate $p$ to find $c$.
 A: Hint:
$$\frac{1}{p(100-p)}=\frac{1}{100}\left(\frac{1}{p}+\frac{1}{100-p}\right)$$
Each term can be integrated separately.
A: Do the integrals: 
$$100\left(-\frac{1}{100}\ln(100 - p) + \frac{1}{100}\ln(p)\right) = t$$
Or, if you want to include the extrema, thence from $p_0$ to $p$ and from $t_0$ to $t$ you get:
$$100\left[\left(-\frac{1}{100}\ln(100 - p) + \frac{1}{100}\ln(p)\right) - \left(-\frac{1}{100}\ln(100 - p_0) + \frac{1}{100}\ln(p_0)\right)\right] = t - t_0$$
Now it's a matter of arranging things and get $p$. Can you do it?
In case, here it is:
$$-\ln(100-p) + \ln(p) + \ln(100 - p_0) - \ln(p_0) = t - t_0$$
Using the properties of logarithm
$$\ln\left(\frac{p(100 - p_0)}{p_0 (100 - p)}\right) = t - t_0$$
Exponentiating
$$\frac{p(100 - p_0)}{p_0(100 - p)} = e^{t - t_0}$$
$$p(100 - p_0) = e^{t - t_0}(p_0(100 - p)) = 100p_0e^{t - t_0} - pp_0e^{t - t_0}$$
Hence
$$p(100 - p_0 + p_0e^{t - t_0}) = 100p_0 e^{t - t_0}$$
$$p(t) = \frac{100p_0 e^{t - t_0}}{100 - p_0 + p_0e^{t - t_0}}$$
A: Solve for $p(t)$ when $\text{n}$ and $\text{m}$ are constants:
$$\text{n}\cdot p'(t)=p(t)\cdot\left(\text{m}-p(t)\right)\Longleftrightarrow\frac{\text{n}\cdot p'(t)}{p(t)\cdot\left(\text{m}-p(t)\right)}=1\Longleftrightarrow\int\frac{\text{n}\cdot p'(t)}{p(t)\cdot\left(\text{m}-p(t)\right)}\space\text{d}t=\int1\space\text{d}t$$
Now use:


*

*$$\int1\space\text{d}t=t+\text{C}$$

*Substitute $u=p(t)$ and $\text{d}u=p'(t)\space\text{d}t$:
$$\int\frac{\text{n}\cdot p'(t)}{p(t)\cdot\left(\text{m}-p(t)\right)}\space\text{d}t=\text{n}\int\frac{1}{u\left(\text{m}-u\right)}\space\text{d}u=\text{n}\left[\int\frac{1}{\text{m}u}\space\text{d}u-\int\frac{1}{\text{m}u-\text{m}^2}\space\text{d}u\right]$$

*$$\int\frac{1}{\text{m}u}\space\text{d}u=\frac{1}{\text{m}}\int\frac{1}{u}\space\text{d}u=\frac{\ln\left|u\right|}{\text{m}}+\text{C}=\frac{\ln\left|p(t)\right|}{\text{m}}+\text{C}$$

*Substitute $s=\text{m}u-\text{m}^2$ and $\text{d}s=\text{m}\space\text{d}u$:
$$\int\frac{1}{\text{m}u-\text{m}^2}\space\text{d}u=\frac{1}{\text{m}}\int\frac{1}{s}\space\text{d}s=\frac{\ln\left|s\right|}{\text{m}}+\text{C}=\frac{\ln\left|\text{m}\cdot p(t)-\text{m}^2\right|}{\text{m}}+\text{C}$$


So, we get:
$$\frac{\text{n}}{\text{m}}\cdot\left(\ln\left|p(t)\right|-\ln\left|\text{m}\cdot p(t)-\text{m}^2\right|\right)=t+\text{C}$$

So, in your problem when $\text{n}=1000$ and $\text{m}=100$, we get:
$$-10\ln\left(\frac{100|p(t)-10|}{|p(t)|}\right)=t+\text{C}$$
