/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 $$ \text { In Problems } 35-38... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 35

$$ \text { In Problems } 35-38, \text { solve each differential equation. } $$ $$ \frac{d y}{d x}=y(1+y) $$

Short Answer

Expert verified
The solution is \( y(x) = \frac{Ce^x}{1 - Ce^x} \), \( C \) is a constant.

Step by step solution

01

Separate Variables

First, we need to rewrite the differential equation \( \frac{dy}{dx} = y(1+y) \) in a form where all \( y \) terms are on one side and all \( x \) terms are on the other side. This can be achieved by rewriting it as \( \frac{1}{y(1+y)} dy = dx \).
02

Integrate Both Sides

Next, we integrate both sides of the equation. The left side becomes \( \int \frac{1}{y(1+y)} \, dy \), and the right side becomes \( \int \ dx \). The left-hand side requires partial fraction decomposition: \[ \frac{1}{y(1+y)} = \frac{A}{y} + \frac{B}{1+y} \] Solve for \( A \) and \( B \): \[ 1 = A(1+y) + By \]Set \( y = 0 \): \( A = 1 \).Set \( y = -1 \): \( B = -1 \). This gives the partial fractions: \[ \int \left( \frac{1}{y} - \frac{1}{1+y} \right) \, dy = \int \, dx \]
03

Evaluate Integrals

For the left side, integrate each part:\[ \int \frac{1}{y} \, dy = \ln |y| \]\[ \int \frac{1}{1+y} \, dy = \ln |1+y| \]Therefore, the integral is:\[ \ln |y| - \ln |1+y| = x + C \], where \( C \) is the constant of integration.
04

Simplify the Result

Combine the logarithms using the property \( \ln a - \ln b = \ln \frac{a}{b} \):\[ \ln \frac{|y|}{|1+y|} = x + C \]To solve for \( y \), remove the natural logarithm by exponentiating both sides:\[ \frac{|y|}{|1+y|} = Ce^x \], where \( C = e^C \).
05

Solve for \( y \) Explicitly

Assume \( y \) is positive, which simplifies the absolute values, and multiply both sides by \( |1+y| \):\[ |y| = Ce^x |1+y| \]Assuming \( y = Ce^x (1+y) \) simplifies to solve for \( y(x) \), considering the scenario \( y > 0 \), results in the explicit solution:\[ y = \frac{Ce^x}{1 - Ce^x} \] for \( Ce^x < 1 \).
06

Final Step: Conclusion of Problem

The solution to the differential equation is given by \( y(x) = \frac{Ce^x}{1 - Ce^x} \), where \( C \) is an arbitrary constant dependent on initial conditions.

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Key Concepts

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

Separation of Variables
Separation of variables is a technique commonly used to solve differential equations. The fundamental idea is to rearrange the equation so that each variable appears on a different side of the equation.
For the given differential equation \( \frac{dy}{dx} = y(1+y) \), this means rewriting it such that all terms involving \( y \) are on one side, and all terms involving \( x \) are on the other side:
  • Start with the equation: \( \frac{dy}{dx} = y(1+y) \).
  • Rearrange to isolate terms: \( \frac{1}{y(1+y)} dy = dx \).
This separates "y " and "x" making it easier to integrate, helping us progress towards finding a solution.
Integration
Integration is the process of finding the integral of a function, which is essentially the reverse operation of differentiation. Once variables are separated, the next step is to integrate both sides.
In the exercise, the left-hand side \( \int \frac{1}{y(1+y)} \, dy \) can be integrated using partial fraction decomposition, while the right-hand side involves a direct integration \( \int \, dx \), which is straightforward:
  • The right side integrates to \( x + C \), where \( C \) is the constant of integration.
  • Partial fractions helps simplify the left side for easier integration.
Integration transforms our separated equation into a solvable form, moving us closer to the solution.
Partial Fraction Decomposition
Partial fraction decomposition is a method for breaking down complex rational expressions into simpler fractions that are easier to integrate.
Consider the expression \( \frac{1}{y(1+y)} \).
  • Express it as \( \frac{A}{y} + \frac{B}{1+y} \).
  • Determine constants \( A \) and \( B \) by equating \( A(1+y) + By = 1 \).
  • Solving yields: \( A = 1 \) and \( B = -1 \).
This decomposition simplifies our integral, allowing us to solve it as \( \int \left( \frac{1}{y} - \frac{1}{1+y} \right) \, dy \), which is straightforward to evaluate.

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Most popular questions from this chapter

Insulin pumps treat patients with type I diabetes by releasing insulin continuously into the fat in the patient's stomach or thigh. We will develop a model for the transport of insulin from the site where it is released by the pump, by treating the fat as a compartment in a single-compartment model. Let's suppose that the pump releases insulin at a constant rate, \(r(r\) is the amount added in one unit of time). (a) Explain why, if insulin is not transported from the site of release, the amount of insulin at the site of release, \(a(t)\), will obey a differential equation: $$ \frac{d a}{d t}=r $$ (b) From the fat, the insulin enters the patient's bloodstream. Suppose that a fraction \(p\) of the insulin present in the patient's fat enters the blood in unit time. Explain why: $$ \frac{d a}{d t}=r-p a $$ (c) Find the equilibrium from the differential equation in part (b) and determine whether this equilibrium is stable or unstable.

Draw the vector field plot of the differential equation. Then, using the given initial conditions, sketch the solutions (i.e., draw a graph showing the dependent variable as a function of the independent variable). \(\frac{d y}{d t}=y^{2}-1\) (a) \(y(0)=-1\), (b) \(y(0)=-1 / 2\), (c) \(y(0)=1 / 2\), (d) \(y(0)=2\).

Find the equilibria of the following differential equations. $$ \frac{d N}{d t}=N \ln N, N>0 $$

Suppose that \(N(t)\) denotes the size of a population at time \(t\). The population evolves according to the logistic equation, but in addition, predation reduces the size of the population so that the rate of change is given by $$ \frac{d N}{d t}=g(N) $$ where $$ g(N)=N\left(1-\frac{N}{50}\right)-\frac{9 N}{5+N} $$ The first term on the right-hand side describes the logistic growth; the second term describes the effect of predation. (a) Make the vector field plot for this differential equation. (b) Find all equilibria of \((8.38)\). (c) Use your vector field plot in (a) to determine the stability of the equilibria you found in (b). (d) Repeat your analysis from part (c) but now use the method of eigenvalues to determine the stability of the equilibria you found in (b).

Draw the vector field plot of the differential equation. Then, using the given initial conditions, sketch the solutions (i.e., draw a graph showing the dependent variable as a function of the independent variable). \(\frac{d y}{d x}=-y-y^{3}\) (a) \(y(0)=0\), (b) \(y(0)=1\), (c) \(y(0)=2\), (d) \(y(0)=3\).
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