# Separate variables and use partial fractions with IVP

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

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

$$x(0)=8$$

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

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

LHS:

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

$$1 = Ax - 5A + 3Bx$$

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.

• Can you state the original problem? – Axion004 Aug 18 at 21:37
• yes I fixed it. Its at the top – K. Gibson Aug 18 at 21:41
• You seem to be using $e^{a+b} = e^a + e^b$. Is that a valid property? – 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 $$\frac{1}{x(x-5)}=\frac{A}{x}+\frac{B}{x-5}=\frac{\frac{-1}{5}}{x}+\frac{\frac{1}{5}}{x-5}=\frac{1}{5}\left(\frac{1}{x-5}-\frac{1}{x}\right)$$

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