separate variables and use partial fractions to solve the initial - value problem. use either the exact…

Explanation:

Step1: Separate variables

We start with the differential equation $\frac{dx}{dt}=-2(x^{2}-1)$. Separate the variables: $\frac{dx}{x^{2}-1}=- 2dt$.

Step2: Decompose into partial - fractions

We know that $\frac{1}{x^{2}-1}=\frac{1}{(x - 1)(x + 1)}=\frac{A}{x - 1}+\frac{B}{x + 1}$. Cross - multiplying gives $1=A(x + 1)+B(x - 1)$. Let $x = 1$, then $A=\frac{1}{2}$, and let $x=-1$, then $B=-\frac{1}{2}$. So $\frac{1}{x^{2}-1}=\frac{1}{2(x - 1)}-\frac{1}{2(x + 1)}$. The differential equation becomes $\left(\frac{1}{2(x - 1)}-\frac{1}{2(x + 1)}\right)dx=-2dt$.

Step3: Integrate both sides

Integrating $\int\left(\frac{1}{2(x - 1)}-\frac{1}{2(x + 1)}\right)dx=\int(-2)dt$. $\frac{1}{2}\ln|x - 1|-\frac{1}{2}\ln|x + 1|=-2t + C$. Using the property of logarithms $\ln a-\ln b=\ln\frac{a}{b}$, we have $\frac{1}{2}\ln\left|\frac{x - 1}{x + 1}\right|=-2t + C$.

Step4: Use the initial condition

Given $x(0)=4$, substitute $t = 0$ and $x = 4$ into $\frac{1}{2}\ln\left|\frac{x - 1}{x + 1}\right|=-2t + C$. $\frac{1}{2}\ln\left(\frac{4 - 1}{4+1}\right)=C$, so $C=\frac{1}{2}\ln\left(\frac{3}{5}\right)$.

Step5: Solve for $x$

We have $\frac{1}{2}\ln\left|\frac{x - 1}{x + 1}\right|=-2t+\frac{1}{2}\ln\left(\frac{3}{5}\right)$. $\ln\left|\frac{x - 1}{x + 1}\right|=-4t+\ln\left(\frac{3}{5}\right)$. $\left|\frac{x - 1}{x + 1}\right|=e^{-4t}\cdot\frac{3}{5}$. Since $x(0) = 4$, $\frac{x - 1}{x + 1}=\frac{3}{5}e^{-4t}$. Cross - multiplying: $5(x - 1)=3e^{-4t}(x + 1)$. $5x-5 = 3e^{-4t}x+3e^{-4t}$. $5x-3e^{-4t}x=3e^{-4t}+5$. $x(5 - 3e^{-4t})=3e^{-4t}+5$. $x(t)=\frac{5 + 3e^{-4t}}{5 - 3e^{-4t}}$.

Answer:

$x(t)=\frac{5 + 3e^{-4t}}{5 - 3e^{-4t}}$