Problem 39 (a) Find the solution to the IVP... [FREE SOLUTION]

91影视

Introductory Differential Equations

Martha L. Abell, James P. Braselton

$Math Studyset 91影视 Explanations$ Math

5 Edition

Chapter 3: Problem 39

(a) Find the solution to the IVP $d y / d t=$ $-r(1-y / A) y, y(0)=y_{0}$, where $r$ and $A$ are positive constants. (b) If $y_{0}A$, show that $\lim _{t \rightarrow \infty} y(t)=\infty$. (d) Show that if $y_{0}>A$, then $y(t)$ has a vertical asymptote at $t=$ $\frac{1}{r} \ln \left(\frac{y_{0}}{y_{0}-A}\right)$.

Short Answer

Expert verified

(a) $y(t) = \frac{A y_0 e^{-rt}}{A-y_0 + y_0 e^{-rt}}$, (b) $\lim_{t \to \infty} y(t) = A$, (c) $\lim_{t \to \infty} y(t) = \infty$, (d) Vertical asymptote at $t = \frac{1}{r} \ln \left( \frac{y_0}{y_0 - A} \right)$.

Step by step solution

- Rewrite the differential equation

Start by rewriting the given differential equation. We have \[\frac{dy}{dt} = -r(1 - \frac{y}{A})y.\]This is a separable differential equation.

- Separate the variables

Separate the variables y and t. We get \[\frac{dy}{y (1 - \frac{y}{A})} = -r dt.\]

- Simplify the left-hand side

The left-hand side can be simplified using partial fractions. We have, \[\frac{1}{y(1 - \frac{y}{A})} = \frac{A}{y(A-y)} = \frac{A}{A(y - y^2/A)}.\]

- Integrate both sides

Integrate both sides of the equation. On the left-hand side, use partial fractions:\[ \int \frac{A}{y(A-y)} dy = \int -r dt.\]Then, this can be separated into:\[ \int \left(\frac{1}{y} + \frac{1}{A-y}\right) dy = -r\int dt.\]Integrate each term separately, which gives us:\[\int \frac{1}{y} dy + \int \frac{1}{A-y} dy = -rt + C.\]

- Solve the integral

Solving the integrals on the left-hand side:\[ \ln|y| - \ln|A-y| = -rt + C.\]

- Simplify and solve for y

Simplify the expression and solve for y:\[ \ln\left(\frac{y}{A-y}\right) = -rt + C.\]Exponentiate both sides to isolate y:\[ \frac{y}{A-y} = Ce^{-rt}.\]From initial conditions, where y(0) = y_0, we find C = \frac{y_0}{A-y_0}.So the solution is\[ \frac{y}{A-y} = \frac{y_0}{A-y_0}e^{-rt}.\]

- Isolate y in the solution

Rearrange to solve for y explicitly:\[ y (A - y_0 e^{-rt}) = A y_0 e^{-rt}.\]Thus,\[ y (A - y) = \, \rightarrow \, y = \frac{A y_0 e^{-rt}}{A - y_0 + y_0 e^{-rt}}.\]

- Evaluate the limit for y_0 < A as t approaches infinity

When $y_0 < A$, evaluate the limit:\[ \lim_{t \to \infty} y(t) = \lim_{t \to \infty} \frac{A y_0 e^{-rt}}{A - y_0 + y_0 e^{-rt}} = A.\]

- Evaluate the limit for y_0 > A as t approaches infinity

If $y_0 > A$, evaluate the limit:\[ \lim_{t \to \infty} y(t) = \lim_{t \to \infty} \frac{A y_0 e^{-rt}}{A - y_0 + y_0 e^{-rt}} = \infty.\]

- Determine the vertical asymptote for y_0 > A

For $y_0 > A$, solve for the vertical asymptote. The solution blows up when the denominator equals zero:\[ A - y_0 + y_0 e^{-rt} = 0.\]Solving for t:\[ y_0 e^{-rt} = y_0 - A.\]Thus,\[ t = \frac{1}{r} \ln \left( \frac{y_0}{y_0 - A} \right).\]

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Separable Differential Equations

In mathematics, a separable differential equation is a type of first-order differential equation that can be written in the form $\frac{dy}{dt} = g(t)h(y)$.
The key feature is that both sides of the equation can be separated, with all the y terms on one side and all the t terms on the other. This makes it possible to solve by integrating both sides.
For instance, in our initial value problem (IVP), we start with the equation $\frac{dy}{dt} = -r(1 - \frac{y}{A})y$.
Separating the variables, we get $\frac{dy}{y(1 - \frac{y}{A})} = -rdt$.
This separation allows us to integrate each side independently, moving towards the solution.

Partial Fraction Decomposition

When solving separable differential equations, it is common to use partial fraction decomposition. This technique simplifies complex fractions into a sum of simpler, more manageable fractions.
In the example given, the fraction $\frac{1}{y(1 - \frac{y}{A})}$ is decomposed as follows: $\frac{A}{y(A-y)} = \frac{A}{A(y - y^2/A)}$.
This allows us to write: $\frac{1}{y(1 - \frac{y}{A})} = \frac{1}{y} + \frac{1}{A-y}$.
By breaking it down this way, we can integrate each simpler fraction separately, which is much easier to handle.

Limit Analysis in Differential Equations

Limit analysis involves determining the behavior of a function as the variable approaches a certain value, often infinity. This is crucial for understanding the long-term behavior of solutions to differential equations.
For our IVP, we need to evaluate the limits for different initial conditions $y_0 < A$ and $y_0 > A$.
When $y_0 < A$, $\text{lim}_{t \rightarrow \text{\textinfty}} y(t) = A$.
Conversely, when $y_0 > A$, the solution diverges to infinity: $\text{lim}_{t \rightarrow \text{\textinfty}} y(t) = \text{\textinfty}$.
This analysis aids in understanding how the solutions behave based on initial conditions.

Vertical Asymptote in Solutions

A vertical asymptote is a line where a function's value grows without bound as it approaches this line.
In the context of differential equations, finding a vertical asymptote can show where the solution becomes infinite.
For solutions with $y_0 > A$, we derive that there is a vertical asymptote when the denominator of $\frac{y}{A - y_0 + y_0 e^{-rt}}$ equals zero.
Solving $A - y_0 + y_0 e^{-rt} = 0$, we find $t = \frac{1}{r} \text{ ln } \frac{y_0}{y_0 - A}$.
This calculation shows exactly when the vertical asymptote occurs.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.