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.

$\frac{dx}{dt}=3x(8 - x), x(0)=13$

$x(t)=\square$

Explanation:

Step1: Separate variables

We start with $\frac{dx}{dt}=3x(8 - x)$. Rearranging gives $\frac{dx}{x(8 - x)}=3dt$.

Step2: Decompose into partial - fractions

We want to express $\frac{1}{x(8 - x)}$ as $\frac{A}{x}+\frac{B}{8 - x}$. Cross - multiplying gives $1 = A(8 - x)+Bx$. Let $x = 0$, then $A=\frac{1}{8}$. Let $x = 8$, then $B=\frac{1}{8}$. So $\frac{1}{x(8 - x)}=\frac{1}{8x}+\frac{1}{8(8 - x)}$, and the differential equation becomes $(\frac{1}{8x}+\frac{1}{8(8 - x)})dx = 3dt$.

Step3: Integrate both sides

Integrating $\int(\frac{1}{8x}+\frac{1}{8(8 - x)})dx=\int3dt$. We know that $\int\frac{1}{8x}dx=\frac{1}{8}\ln|x|$ and $\int\frac{1}{8(8 - x)}dx=-\frac{1}{8}\ln|8 - x|$. So $\frac{1}{8}\ln|\frac{x}{8 - x}| = 3t + C$.

Step4: Use the initial condition

Given $x(0)=13$, substitute $t = 0$ and $x = 13$ into $\frac{1}{8}\ln|\frac{x}{8 - x}| = 3t + C$. We get $\frac{1}{8}\ln|\frac{13}{8 - 13}|=C$, i.e., $C=\frac{1}{8}\ln|\frac{13}{- 5}|=\frac{1}{8}\ln\frac{13}{5}$.

Step5: Solve for $x$

From $\frac{1}{8}\ln|\frac{x}{8 - x}| = 3t+\frac{1}{8}\ln\frac{13}{5}$, we have $\ln|\frac{x}{8 - x}| = 24t+\ln\frac{13}{5}$. Then $\frac{x}{8 - x}=\frac{13}{5}e^{24t}$. Cross - multiplying gives $5x=(8 - x)\times13e^{24t}$, $5x = 104e^{24t}-13xe^{24t}$, $x(5 + 13e^{24t})=104e^{24t}$, and $x(t)=\frac{104e^{24t}}{5 + 13e^{24t}}$.

Answer:

$x(t)=\frac{104e^{24t}}{5 + 13e^{24t}}$