Partial fractions ODE first order

Question

Given the problem: $\frac{dx}{dt} = 3x(x-5)$ The answer is supposedly:

$\dfrac{40}{8-3e^{-15t}}$ with $x(0) = 8$

first off you can't have a negative power in the denominator it should just be $e^{15t}$.

I know for a fact I did my partial fractions correctly:

$1 = \dfrac{A}{3x} + \dfrac{B}{x-5}$ translates to: $Ax-5A+15xB$

At $x = 0: A = -\frac{1}{5}$

At $x = 5: B = \frac{1}{15}$

Of course this comes to:

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

which becomes:

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

multiply by $-15$ to give:

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

multiplying by $e$:

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

$2x = Ce^{-15t}-5$

$x(t) = \dfrac{2}{Ce^{15t}} - 5$

$8 = \dfrac{2}{C}-5 \to C = \dfrac{2}{13}$

Now I am thinking there is either a mistake in the book or they "simplified" it somehow

Answer 1

$$ 1 = \frac{A}{3x} + \frac{B}{x-5} $$ if and only if $$ (3x)(x-5) = A(x-5) + B(3x) \text{,} $$ with $x \not\in \{0,5\}$. How do we get "$15xB$"?

Answer 2

Before giving you a complete solution I want to point out the mistakes in your solution.

1. Your partial fractions are correct.
2. As mentioned in the comments $\int \frac{1}{ax} dx \ne \ln | a x |$. Instead we have $$ \int \frac{1}{ax} dx = \frac{1}{a} \int \frac{1}{x} dx = \frac{1}{a} \ln | x| $$ for $a \ne 0$.
3. We have $e^{\ln(a) - \ln(b)} \ne a - b$. This is because $\ln(a) - \ln(b) = \ln \left(\frac{a}{b}\right)$ and therefore $$ e^{\ln(a) - \ln(b)} = \frac{a}{b}. $$ Also, as mentioned in the comments, raising both sides to the power of $e$ is called exponentiating, not multiplying by $e$.

Here's how I would do it:

---

Using separation of variables we write $$ \frac{x'(t)}{x(t)(x(t) - 5)} = 3 $$ and then integrate $$ \int_{y_0}^{y} \frac{x'(t)}{x(t)(x(t) - 5)} dt = \int_{y_0}^{y} 3 dt = 3(y - y_0) $$ Now do partial fractions: $$ \frac{1}{x(x - 5)} = \frac{A}{x} + \frac{B}{x - 5} \implies 1 = A(x - 5) + B x $$ Now plugging in $x = 0$ we have $$ 1 = -5 A \implies A = - \frac{1}{5}. $$ Plugging in $x = 5$ we have $$ 1 = 5B \implies B = \frac{1}{5}. $$ Therefore we get $$ \frac{1}{x(x - 5)} = \frac{1}{5(x - 5)} -\frac{1}{5x}. $$ This now gives by substitution \begin{align} \int_{y_0}^{y} \frac{x'(t)}{x(t) ( x(t) - 5)} dt & = \int_{x(y_0)}^{x(y)} \frac{1}{r(r - 5)} dr = \frac{1}{5}\int_{x(y_0)}^{x(y)} \frac{1}{r - 5} -\frac{1}{r} dr \\ & = \frac{1}{5} \left( \ln \left| \frac{x(y) - 5}{x(y_0)} \right| - \ln\left| \frac{x(y)}{x(y_0)} \right| \right) = \frac{1}{5} \ln\left| \frac{(x(y) - 5)x(y_0)}{x(y) x(y_0)} \right| \\ & = \frac{1}{5} \ln\left| \frac{x(y) - 5}{x(y)} \right| \end{align} Therefore, we have \begin{align} \frac{1}{5} \ln\left| \frac{x(y) - 5}{x(y)} \right| = 3(y - y_0) \implies & \ln\left| \frac{x(y) - 5}{x(y)} \right| = 15(y - y_0) \\ \implies & \frac{x(y) - 5}{x(y)} = e^{15(y - y_0)} \\ \implies & x(y) = \frac{5}{1 - e^{15(y - y_0)}} \end{align} Now we have $x(0) = 8$. Plugging in we have $$ x(0) = \frac{5}{1 - e^{-15y_0}} \overset{!}{=} 8 \implies e^{-15y_0} = \frac{3}{8} \implies y_0 = \frac{\ln\left(\frac{8}{3}\right)}{15} $$ Plugging in we obtain $$ x(y) = \frac{5}{1 - e^{15(y - \frac{\ln\left(\frac{8}{3}\right)}{15}}} = \frac{5}{1 - e^{15y - \ln\left(\frac{8}{3}\right)}} = \frac{5}{1 - \frac{e^{15y}}{\frac{8}{3}}} = \frac{5}{1 - \frac{3e^{15y}}{8}} = \frac{40}{8 - 3e^{15y}}, $$ as required.

---

Also notice that by dividing by $x(x-5)$ we exclude the constant solutions $x \equiv 0$ and $x \equiv 5$. Both solve the differential equation but not the initial value problem because they don't fulfil $x(0) = 8$.

Answer 3

HINT \begin{align*} \frac{1}{3x(x-5)} = \frac{1}{3}\times\frac{1}{x(x-5)} = \frac{1}{15}\times\frac{x - (x-5)}{x(x-5)} = \frac{1}{15(x-5)} - \frac{1}{15x} \end{align*}

Answer 4

Another way would be to explicitly calculate the coefficients of the partial fractions.

We are given that

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

which rearranges to (observing that $x\equiv 0$ and $x \equiv 5$ are solutions to the ODE but not the IVP)

$$\frac{1}{x(x-5)}dx=3dt \implies-\frac{1}{5}\Big(\frac{(x-5)-x}{x(x-5)}\Big)dx=3dt\implies -\frac{1}{5}\Big(\frac{1}{x}-\frac{1}{x-5}\Big)dx=3dt$$

so that

$$-\frac{1}{5}\ln|x|+\frac{1}{5}\ln|x-5|=3t+C \implies\ln\Big(\frac{|x-5|}{|x|}\Big)=15t+C$$

which reduces to

$$x=\frac{5}{1-Ce^{15t}}$$

Evaluating the initial condition of $x(0)=8$ forms

$$8=\frac{5}{1-C}\implies C=\frac{3}{8}$$

where

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

Answer 5

You can, as alternative solution method, avoid using a partial fraction decomposition by treating the ODE as a Bernoulli DE $$ x'=3x^2-15x $$ so that by setting $u=x^{-1}$ you get $$ u'=-x^{-2}x'=15x^{-1}-3=15u-3 $$ which now is a linear first order DE in $u$ and additionally separable. Integration leads to $$ 5u(t)-1=e^{15t}(5u(0)-1)\implies 5\frac{x(0)}{x(t)}=x(0)+e^{15t}(5-x(0)) \\~\\ x(t)=\frac{5x(0)}{x(0)+e^{15t}(5-x(0))}. $$