question 3 of 5 > find the solution of dy/dt = 6y(5 - y), y(0) = 25. (express numbers in exact form. use…

Explanation:

Step1: Separate variables

Rewrite the differential equation $\frac{dy}{dt}=6y(5 - y)$ as $\frac{dy}{y(5 - y)}=6dt$.

Step2: Decompose the fraction

Use partial - fraction decomposition on $\frac{1}{y(5 - y)}$. We know that $\frac{1}{y(5 - y)}=\frac{A}{y}+\frac{B}{5 - y}$. Solving for $A$ and $B$ gives $A = \frac{1}{5}$ and $B=\frac{1}{5}$. So $\frac{1}{y(5 - y)}=\frac{1}{5y}+\frac{1}{5(5 - y)}$, and the equation becomes $(\frac{1}{5y}+\frac{1}{5(5 - y)})dy = 6dt$.

Step3: Integrate both sides

Integrate: $\int(\frac{1}{5y}+\frac{1}{5(5 - y)})dy=\int6dt$. The left - hand side integral is $\frac{1}{5}\int\frac{1}{y}dy+\frac{1}{5}\int\frac{1}{5 - y}dy$. Let $u = 5 - y$ for the second integral, then $du=-dy$. We get $\frac{1}{5}\ln|y|-\frac{1}{5}\ln|5 - y| = 6t + C$. Using the property of logarithms $\ln a-\ln b=\ln\frac{a}{b}$, we have $\frac{1}{5}\ln|\frac{y}{5 - y}| = 6t + C$.

Step4: Use the initial condition

Given $y(0)=25$, substitute $t = 0$ and $y = 25$ into $\frac{1}{5}\ln|\frac{y}{5 - y}| = 6t + C$. $\frac{1}{5}\ln|\frac{25}{5 - 25}|=C$, $\frac{1}{5}\ln|\frac{25}{- 20}|=C$, $\frac{1}{5}\ln(\frac{5}{4})=C$.

Step5: Solve for $y$

From $\frac{1}{5}\ln|\frac{y}{5 - y}| = 6t+\frac{1}{5}\ln(\frac{5}{4})$, we have $\ln|\frac{y}{5 - y}| = 30t+\ln(\frac{5}{4})$. Exponentiating both sides gives $\frac{y}{5 - y}=\frac{5}{4}e^{30t}$. Cross - multiply: $4y = 25e^{30t}-5ye^{30t}$. Collect terms with $y$: $4y + 5ye^{30t}=25e^{30t}$. Factor out $y$: $y(4 + 5e^{30t})=25e^{30t}$. $y=\frac{25e^{30t}}{4 + 5e^{30t}}=\frac{25}{5 - 4e^{-30t}}$ (after multiplying numerator and denominator by $e^{- 30t}$).

Answer:

$y=\frac{25}{5 - 4e^{-30t}}$