Question

(a) Find the biomass in the later year with the initial condition\(y(0) = 2 \times {10^7}\;{\rm{kg}}\). The differential equation for the fishery is\(\frac{{dy}}{{dt}} = ky(M - y)\).

(b) Find the time at which the biomass touches the\(4 \times {10^7}\;{\rm{kg mark }}\).

Short answer

(a) The biomass in the later year with the initial condition \(y(0) = 2 \times {10^7}kg\)is approximately \(2 \times {10^7}\;{\rm{kg}}.\)

(b) The time at which the biomass touches the \(4 \times {10^7}\;{\rm{kg}}\) mark is \(123787300.132{\rm{ }}\) years.

Step-by-step solution

Solve the differential equation to obtain the equation.

It is given that the differential equation for the fishery is \(\frac{{dy}}{{dt}} = ky(M - y)\) and the initial condition is \(y(0) = 2 \times {10^7}\;{\rm{kg}}\).

Solve the differential equation as follows.

\(\begin{aligned}frac{{dy}}{{y(M - y)}} &= kdt\\\frac{{dy}}{{My}} - \frac{{dy}}{{M(M - y)}} &= kdt\\\frac{{\ln |y|}}{M} - \frac{{\ln |M - y|}}{M} &= kt + C\\\frac{1}{M}\ln \frac{y}{{M - y}} &= kt + C\end{aligned}\)

On further simplification,

\(\begin{aligned}\ln \frac{y}{{M - y}} &= kt + C\\\frac{y}{{M - y}}& = {e^{kt + C}}\\\frac{y}{{M - y}}& = C{e^{kt}}\\\frac{{M - y}}{y}& = C{e^{ - kt}}\end{aligned}\)

Thus, the equation is obtained as,

\(\begin{aligned}\frac{M}{y} - 1 &=C{e^{ - kt}}\\1 + C{e^{ - kt}} &=\frac{M}{y}\\y &=\frac{M}{{1 + C{e^{ - kt}}}}\end{aligned}\)

Obtain the value of \(C\).

As \(y(0) = 2 \times {10^7}\;{\rm{kg}}\), the value of \(C\) becomes,

Thus, the solution of the differential equation is\(y(t) = \frac{{8 \times {{10}^7}}}{{1 + 3{e^{ - 8.875 \times {{10}^{ - 9}}}}}}{\rm{ as }}k = 8.875 \times {10^{ - 9}}\)per year,

At\(t = 1\), the biomass becomes\(y(1) = \frac{{8 \times {{10}^7}}}{{1 + 3{e^{ - 8.875 \times {{10}^{ - 9}}(1)}}}} \approx 2 \times {10^7}\;{\rm{kg}}\).

Therefore, the biomass in the later year with the initial condition \(y(0) = 2 \times {10^7}kg\) is approximately \(2 \times {10^7}\;{\rm{kg}}.\)

Obtain the value of \(t\).

From part (a), the solution of the differential equation is \(y(t) = \frac{{8 \times {{10}^7}}}{{1 + 3{e^{ - 8.875 \times {{10}^{ - 9}}}}}}{\rm{ as }}k = 8.875 \times {10^{ - 9}}\) per year.

Now obtain the value of\(t\)for which the biomass is\(4 \times {10^7}\;{\rm{kg}}\).

\(\begin{aligned}4 \times {10^7} &= \frac{{8 \times {{10}^7}}}{{1 + 3{e^{ - 8.875 \times {{10}^{ - 9}}t}}}}\\1 + 3{e^{ - 8.875 \times {{10}^{ - 9}}t}} &= \frac{{8 \times {{10}^7}}}{{4 \times {{10}^7}}}\\3{e^{ - 8.875 \times {{10}^{ - 9}}t}} &= 2 - 1\\{e^{ - 8.875 \times {{10}^{ - 9}}t}} &= \frac{1}{3}\end{aligned}\)

Thus, the value of\(t\)becomes,

\(\begin{aligned} - 8.875 \times {10^{ - 9}}t &= \ln \frac{1}{3}\\8.875 \times {10^{ - 9}}t &= \ln 3\\t &= \frac{{\ln 3}}{{8.875 \times {{10}^{ - 9}}}}\\t \approx 123787300.132{\rm{ years}}\end{aligned}\)

Therefore, the time at which the biomass touches the \(4 \times {10^7}\;{\rm{kg}}\) mark is \(123787300.132{\rm{ }}\) years.