# Help solving differential equation?

Solve the differential equation:

$$\dfrac{dE}{dt}=f(E)=\dfrac{E(9-E)(E-4)}{24}$$

• let E(0)=7 thousand this represents a population of elk (in thousands) in a national park – Lil Feb 27 '15 at 3:13
• $f(E)$ was provided, $f(E) = E(9-E)(E-4)/24$ just as you wrote. – nullUser Feb 27 '15 at 3:23
• I apologize. I made a typo. – Lil Feb 27 '15 at 3:25
• right now I have a very basic understanding of differential equations. So I know how to solve equations using variation of parameters, separation of variables, partial fractions, and integration factors. I also learned how to find the slope field – Lil Feb 27 '15 at 3:27

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 $E$.

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

$$\frac{ E^{1/36} ( E-9)^{1/45}} {(E-4)^{1/20}} = \tilde C \exp \left ( - \frac{t}{24} \right )$$

$$\frac{E^{900} (E-9)^{720}}{ (E-4)^{1620}} = \hat C \exp ( - 1350 t)$$