Solve $\frac{dx}{dt} = 3x(x-5)$

This is all done in terms of $x(t)$:


$\frac{dx}{dt} = 3x(x-5)$

$\int \frac{dx}{3x(x-5)} = \int dt$


$\frac{A}{3x} + \frac{B}{x-5}$

$1 = Ax - 5A + 3Bx$

let $x = 5$ then $B = \frac{1}{15}$

let $x = 0$ then $A = -\frac{1}{5}$

$\int -\frac{\frac{1}{5}}{3x} + \int \frac{\frac{1}{15}}{x-5}$

$-\frac{1}{5}ln \vert 3x \vert + \frac{1}{15}ln \vert x-5 \vert = t + C$

multiply through by by $-15$

$3ln \vert 3x \vert - ln\vert x - 5 \vert = -15t + C$

multiplying through by $e$ gives

$3x^3 -x-5=e^{-15t} + C$

Plugging in IVP:

$1 + C = 3(512)-8-6=1523$

Somehow the answer is $x(t) = \frac{40}{8-e^{-15t}}$

this is asinine.

  • $\begingroup$ Can you state the original problem? $\endgroup$ – Axion004 Aug 18 at 21:37
  • $\begingroup$ yes I fixed it. Its at the top $\endgroup$ – K. Gibson Aug 18 at 21:41
  • 1
    $\begingroup$ You seem to be using $e^{a+b} = e^a + e^b$. Is that a valid property? $\endgroup$ – Ted Shifrin Aug 18 at 22:28

$$\frac{dx}{dt}=3x(x-5)\implies\frac{dx}{x(x-5)}=3dt\implies$$ $$\int\frac{dx}{x(x-5)}=\int3dt\implies\int\frac{1}{5}\left(\frac{1}{x-5}-\frac{1}{x}\right)dx=3t+c\implies$$ $$\frac{1}{5}\ln|x-5|-\frac{1}{5}\ln|x|=\frac{1}{5}\ln\left|\frac{x-5}{x}\right|=3t+c$$

Taking the exponential of both sides, $$\left(\frac{x-5}{x}\right)^{\frac{1}{5}}=e^{3t+c}=e^{3t}e^c=ke^{3t}$$

Thus $$\frac{x-5}{x}=k^5e^{15t}\implies x-5=xk^5e^{15t}\implies x\left(1-k^5e^{15t}\right)=5$$ $$\implies x(t)=\frac{5}{1-k^5e^{15t}}$$

With $x(0)=8,$ $$x(0)=\frac{5}{1-k^5}=8\implies5=8-8k^5\implies\frac{3}{8}=k^5$$

Thus $$x(t)=\frac{5}{1-\frac{3}{8}e^{15t}}\cdot\frac{8}{8}=\frac{40}{8-3e^{15t}}$$

Addendum: $$\frac{1}{x(x-5)}=\frac{A}{x}+\frac{B}{x-5}\implies $$ $$x(x-5)\cdot\frac{1}{x(x-5)}=x(x-5)\cdot\left(\frac{A}{x}+\frac{B}{x-5}\right)\implies$$ $$1=\frac{Ax(x-5)}{x}+\frac{Bx(x-5)}{x-5}\implies$$ $$1=A(x-5)+Bx\implies$$ $$1=x(A+B)-5A\tag1$$

Since there are no variables on the LHS of $(1)$, the coefficient of $x$ of the RHS, namely $A+B$, must be $0$. On the other hand, there are constants on the left and the right of $(1)$. Thus we again equate coeffients so that $1=-5A$. Hence we have two conditions $$A+B=0\text{ and } 1=-5A$$ The latter equation implies $A=-\frac{1}{5}$ so that when substituting this value of $A$ into the equation $A+B=0$ we find that $B=\frac{1}{5}$. Finally we find $$\frac{1}{x(x-5)}=\frac{A}{x}+\frac{B}{x-5}=\frac{\frac{-1}{5}}{x}+\frac{\frac{1}{5}}{x-5}=\frac{1}{5}\left(\frac{1}{x-5}-\frac{1}{x}\right)$$

  • $\begingroup$ where did you get $\frac{1}{5}$ $\endgroup$ – K. Gibson Aug 18 at 22:05
  • $\begingroup$ and subtraction sign $\endgroup$ – K. Gibson Aug 18 at 22:06
  • $\begingroup$ See my addendum. $\endgroup$ – coreyman317 Aug 18 at 22:21
  • $\begingroup$ If this answer was sufficient, please consider 'accepting' it by clicking the green checkmark next to the upvotes. Thank you. $\endgroup$ – coreyman317 Aug 22 at 20:34

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