Solve the differential equation:


I'm not sure how to go about this? Do I do this through partial fractions?

  • $\begingroup$ let E(0)=7 thousand this represents a population of elk (in thousands) in a national park $\endgroup$ – Lil Feb 27 '15 at 3:13
  • $\begingroup$ $f(E)$ was provided, $f(E) = E(9-E)(E-4)/24$ just as you wrote. $\endgroup$ – nullUser Feb 27 '15 at 3:23
  • $\begingroup$ I apologize. I made a typo. $\endgroup$ – Lil Feb 27 '15 at 3:25
  • $\begingroup$ 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 $\endgroup$ – Lil Feb 27 '15 at 3:27

The equation $\dfrac{dE}{dt}=\dfrac{E(9−E)(E−4)}{24}$ is separable:



$\int \dfrac{24dE}{E(9−E)(E−4)}= \int dt$



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

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

Simplifying (though adding extraneous solutions)

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


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