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. $\frac{dx}{dt}=5x(x - 7),x(0)=1$ $x(t)=square$

Explanation:

Step1: Separate variables

Separate the variables in the differential equation $\frac{dx}{dt}=5x(x - 7)$. We get $\frac{dx}{x(x - 7)}=5dt$.

Step2: Decompose into partial - fractions

We decompose $\frac{1}{x(x - 7)}$ into partial fractions. Let $\frac{1}{x(x - 7)}=\frac{A}{x}+\frac{B}{x - 7}$. Then $1=A(x - 7)+Bx$. Setting $x = 0$, we find $A=-\frac{1}{7}$. Setting $x = 7$, we find $B=\frac{1}{7}$. So $\frac{1}{x(x - 7)}=\frac{-1/7}{x}+\frac{1/7}{x - 7}$, and the differential equation becomes $(-\frac{1}{7x}+\frac{1}{7(x - 7)})dx = 5dt$.

Step3: Integrate both sides

Integrate both sides: $\int(-\frac{1}{7x}+\frac{1}{7(x - 7)})dx=\int5dt$.
The left - hand side integral: $-\frac{1}{7}\int\frac{1}{x}dx+\frac{1}{7}\int\frac{1}{x - 7}dx=-\frac{1}{7}\ln|x|+\frac{1}{7}\ln|x - 7|+C_1$.
The right - hand side integral: $\int5dt = 5t+C_2$.
Using the property of logarithms $\ln a-\ln b=\ln\frac{a}{b}$, we have $\frac{1}{7}\ln|\frac{x - 7}{x}|=5t + C$.

Step4: Use the initial condition

Given $x(0)=1$. Substitute $t = 0$ and $x = 1$ into $\frac{1}{7}\ln|\frac{x - 7}{x}|=5t + C$.
$\frac{1}{7}\ln|\frac{1 - 7}{1}|=C$, so $C=\frac{1}{7}\ln6$.

Step5: Solve for $x$

We have $\frac{1}{7}\ln|\frac{x - 7}{x}|=5t+\frac{1}{7}\ln6$.
$\ln|\frac{x - 7}{x}|=35t+\ln6$.
$\frac{x - 7}{x}=6e^{35t}$.
$x - 7 = 6xe^{35t}$.
$x-6xe^{35t}=7$.
$x(1 - 6e^{35t})=7$.
$x(t)=\frac{7}{1 - 6e^{35t}}$.

Answer:

$x(t)=\frac{7}{1 - 6e^{35t}}$