Let y (t) be a solution of dot y = {1} / {6} y(1 - {y}/ {6}) such that y (0) = 12. Determine...

Question:

Let {eq}y (t) {/eq} be a solution of {eq}\dot y = \frac {1} {6} y\bigg(1 - \frac {y} {6}\bigg) {/eq} such that {eq}y (0) = 12 {/eq}. Determine {eq}\displaystyle \lim_{t\ \to\ \infty} y (t) {/eq} without finding {eq}y (t) {/eq} explicitly.

Anti-derivative:

Let {eq}f'\left( x \right) {/eq} be a function. Integrate the function {eq}f'\left( x \right) {/eq} as {eq}\int {f'\left( x \right)} = f\left( x \right) + c {/eq}. Here, c is a integration constant and {eq}f\left( x \right) {/eq} is called the anti-derivative of the function. So, the anti-derivative of a function can be found by integrating the function.

Answer and Explanation:

Given

  • The function is given by {eq}y' = \dfrac{1}{6}y\left( {1 - \dfrac{y}{6}} \right) {/eq} with initial condition {eq}y\left( 0 \right) = 12 {/eq}.

Find the value of {eq}\mathop {\lim }\limits_{t \to \infty } y\left( t \right) {/eq}.

Rewrite the given function.

{eq}\begin{align*} \dfrac{{dy}}{{dt}} &= \dfrac{1}{6}y\left( {1 - \dfrac{y}{6}} \right)\\ \dfrac{{dy}}{{y\left( {1 - y/6} \right)}} &= \dfrac{1}{6}dt \end{align*} {/eq}

Integrate both sides of the above equation.

{eq}\int {\dfrac{{dy}}{{y\left( {1 - y/6} \right)}}} = \int {\dfrac{1}{6}dt} {/eq}

Apply partial decomposition method to solve the above integral.

{eq}\begin{align*} \int {\dfrac{{dy}}{y} + \dfrac{1}{6}\int {\dfrac{{6dy}}{{6 - y}}} } &= \int {\dfrac{1}{6}dt} \\ \int {\dfrac{{dy}}{y} + \int {\dfrac{{dy}}{{6 - y}}} } &= \int {\dfrac{1}{6}dt} \\ \int {\dfrac{{dy}}{y} - \int {\dfrac{{dy}}{{y - 6}}} } &= \int {\dfrac{1}{6}dt} \\ \ln y - \ln \left( {y - 6} \right) &= \dfrac{t}{6} + c\\ \ln \left( {\dfrac{y}{{y - 6}}} \right) &= \dfrac{t}{6} + c\\ \dfrac{y}{{y - 6}} &= {e^{\dfrac{t}{6} + c}}\\ \dfrac{y}{{y - 6}} &= {e^{\dfrac{t}{6}}}{e^c}\\ \dfrac{y}{{y - 6}} &= C{e^{\dfrac{t}{6}}} \end{align*} {/eq}

Here, {eq}{e^c} = C {/eq}.

Solve the above equation in terms of {eq}y\left( t \right) {/eq}.

{eq}\begin{align*} y &= \left( {y - 6} \right)C{e^{\dfrac{t}{6}}}\\ &= yC{e^{\dfrac{t}{6}}} - 6C{e^{\dfrac{t}{6}}}\\ y\left( {1 - C{e^{\dfrac{t}{6}}}} \right) &= - 6C{e^{\dfrac{t}{6}}}\\ y\left( t \right) &= \dfrac{{ - 6C{e^{\dfrac{t}{6}}}}}{{1 - C{e^{\dfrac{t}{6}}}}} \ldots \left( 1 \right) \end{align*} {/eq}

As {eq}y\left( 0 \right) = 12 {/eq}. So, substitute 0 for {eq}t {/eq} and 12 for {eq}y\left( t \right) {/eq} in above equation to find the value of {eq}C {/eq}.

{eq}\begin{align*} 12 &= \dfrac{{ - 6C{e^0}}}{{1 - C{e^0}}}\\ 12 &= \dfrac{{ - 6C}}{{1 - C}}\\ 12 - 12C &= - 6C\\ 12C - 6C &= 12\\ C &= 2 \end{align*} {/eq}

Substitute 2 for {eq}C {/eq} in the equation (1) then,

{eq}y\left( t \right) = \dfrac{{ - 12{e^{\dfrac{t}{6}}}}}{{1 - 2{e^{\dfrac{t}{6}}}}} {/eq}

Apply the limit as {eq}t {/eq} approaches infinity on the above equation.

{eq}\begin{align*} \mathop {\lim }\limits_{t \to \infty } y\left( t \right) &= \mathop {\lim }\limits_{t \to \infty } \dfrac{{ - 12{e^{\dfrac{t}{6}}}}}{{1 - 2{e^{\dfrac{t}{6}}}}}\\ &= \mathop {\lim }\limits_{t \to \infty } \dfrac{{ - 12{e^{\dfrac{t}{6}}}}}{{{e^{\dfrac{t}{6}}}\left( {1/{e^{\dfrac{t}{6}}} - 2} \right)}}\\ &= \mathop {\lim }\limits_{t \to \infty } \dfrac{{ - 12}}{{\left( {1/{e^{\dfrac{t}{6}}} - 2} \right)}}\\ &= \dfrac{{ - 12}}{{1/{e^\infty } - 2}}\\ &= \dfrac{{ - 12}}{{0 - 2}}\\ &= 6 \end{align*} {/eq}

Thus, the value of{eq}\mathop {\lim }\limits_{t \to \infty } y\left( t \right) {/eq} is 6.


Learn more about this topic:

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Indefinite Integrals as Anti Derivatives

from Math 104: Calculus

Chapter 12 / Lesson 11
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