Question

separate variables and use partial fractions to solve the initial - value problem. use either the exact solution or a computer - generated slope field to sketch the graphs of several solutions of the given differential equation, and highlight the indicated particular solution. $\frac{dx}{dt}=3x(x - 8),x(0)=3$ $x(t)=square$

Explanation:

Step1: Separate variables

We have $\frac{dx}{dt}=3x(x - 8)$. Separate variables: $\frac{dx}{x(x - 8)}=3dt$.

Step2: Decompose into partial - fractions

We decompose $\frac{1}{x(x - 8)}$ into partial fractions. Let $\frac{1}{x(x - 8)}=\frac{A}{x}+\frac{B}{x - 8}$. Then $1=A(x - 8)+Bx$. Setting $x = 0$, we get $A=-\frac{1}{8}$. Setting $x = 8$, we get $B=\frac{1}{8}$. So $\frac{1}{x(x - 8)}=\frac{-1/8}{x}+\frac{1/8}{x - 8}$, and the differential equation becomes $(-\frac{1}{8x}+\frac{1}{8(x - 8)})dx=3dt$.

Step3: Integrate both sides

Integrating gives $-\frac{1}{8}\int\frac{1}{x}dx+\frac{1}{8}\int\frac{1}{x - 8}dx=\int3dt$.
$-\frac{1}{8}\ln|x|+\frac{1}{8}\ln|x - 8| = 3t+C$.
Using the property of logarithms $\ln a-\ln b=\ln\frac{a}{b}$, we have $\frac{1}{8}\ln|\frac{x - 8}{x}|=3t+C$.

Step4: Use the initial condition

Given $x(0)=3$. Substitute $t = 0$ and $x = 3$ into $\frac{1}{8}\ln|\frac{x - 8}{x}|=3t+C$.
$\frac{1}{8}\ln|\frac{3 - 8}{3}|=C$, so $C=\frac{1}{8}\ln|\frac{-5}{3}|=\frac{1}{8}\ln\frac{5}{3}$.

Step5: Solve for $x$

We have $\frac{1}{8}\ln|\frac{x - 8}{x}|=3t+\frac{1}{8}\ln\frac{5}{3}$.
$\ln|\frac{x - 8}{x}|=24t+\ln\frac{5}{3}$.
$\frac{x - 8}{x}=e^{24t}\cdot\frac{5}{3}$.
$x - 8=\frac{5}{3}xe^{24t}$.
$x-\frac{5}{3}xe^{24t}=8$.
$x(1-\frac{5}{3}e^{24t})=8$.
$x(t)=\frac{24}{3 - 5e^{24t}}$.

Answer:

$x(t)=\frac{24}{3 - 5e^{24t}}$