Differential equation $x'=11x -x^2 -24$

Question

Getting stuck finding $x(t)$ on the differential equation: $dx/dt = 11x -x^2 -24$ with $x(0)=5$.

So my work so far is: $dx/(11x-x^2-24) = dt$

Using partial fractions $A(x-3) + B(x-8) = 1$, so $A = 1/5$, $B=-1/5$ This gives me the integral: $\int(1/(x-8)-1/(x-3)) dx = 5dt$

$t+c = \ln(x-8)-\ln(x-3)$

$t+c = \ln((x-8)/(x-3))$

$Ce^t = (x-8)/(x-3)$

I know $C$ is $-3/2$.

How do I get this in terms of $x(t)$? I keep getting stuck on this step when solving differential equations and I'm hoping can help me with this specific step. Thanks!

Answer 1

Note that you seemed to have mixed up your signs; partial fraction decomposition gives us that: $$ \frac{1}{11x-x^2-24} = \frac{-1/5}{x-8} + \frac{1/5}{x-3} $$

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Cross-multiply, bring the $x$ terms together, factor out the $x$, then divide through: \begin{align*} \frac{3e^t}{2} &= \frac{x - 8}{x - 3} \\ 3xe^t - 9e^t &= 2x - 16 \\ 3xe^t - 2x &= 9e^t - 16 \\ x(3e^t - 2) &= 9e^t - 16 \\ x(t) &= \frac{9e^t - 16}{3e^t - 2} \\ \end{align*}