Help solving differential equation? Solve the differential equation: 
$$\dfrac{dE}{dt}=f(E)=\dfrac{E(9-E)(E-4)}{24}$$
I'm not sure how to go about this? Do I do this through partial fractions? 
 A: The equation $\dfrac{dE}{dt}=\dfrac{E(9−E)(E−4)}{24}$ is separable:
$\dfrac{24dE}{E(9−E)(E−4)}=dt$
Integrate
$\int \dfrac{24dE}{E(9−E)(E−4)}= \int dt$
then
$-\dfrac{2}{15}(-9\log(4-E)+4\log(9-E)+5\log(E))+c=t$
only clears $E$.
A: You do use partial fractions, which will give you
$$\frac{1}{y(y-9)(y-4)} = \frac{1}{36 y} + \frac{1}{45(y-9)} -\frac{1}{20(y-4)} $$ 
Thus, since the equation is separable, we have
$$ E' = \frac{ E(E-9)(4-E)}{24} \iff \int \frac{ dE }{E(E-9)(E-4) } = -\frac{1}{24} \int dt $$
Using the partial fractions now, we see the integrals are
$$ \frac{1}{36}\ln |E| + \frac{1}{45}\ln | E - 9 | - \frac{1}{20} \ln |y-4| =- \frac{t}{24} +C \quad \text{where} \quad C \in \mathbb{R} $$
Thus we obtain
$$ \ln E^{1/36} + \ln ( E - 9)^{1/45} - \ln (E-4)^{1/20} = -\frac{t}{24} + C $$
Exponentiating both sides now we see
$$ \frac{ E^{1/36} ( E-9)^{1/45}} {(E-4)^{1/20}} = \tilde C \exp \left ( - \frac{t}{24} \right )$$
Simplifying (though adding extraneous solutions)
$$ \frac{E^{900} (E-9)^{720}}{ (E-4)^{1620}} = \hat C \exp ( - 1350 t) $$
