/*! 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 31 Use the partial-fraction method ... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 31

Use the partial-fraction method to solve $$ \frac{d y}{d x}=y(y-2) $$ where \(y(0)=1\)

Short Answer

Expert verified
The solution is \( y = \frac{2e^{2x}}{1-e^{2x}} \).

Step by step solution

01

Identify the Differential Equation

The given differential equation is \( \frac{d y}{d x} = y(y-2) \). This is a separable differential equation because the right-hand side depends only on \( y \).
02

Set Up Separate Variables

We rearrange the equation to separate variables: \( \frac{d y}{y(y-2)} = dx \). This allows us to integrate each side with respect to their own variables.
03

Partial Fraction Decomposition

Use partial fraction decomposition to express \( \frac{1}{y(y-2)} \). Assume \( \frac{1}{y(y-2)} = \frac{A}{y} + \frac{B}{y-2} \). Solving gives \( A = \frac{1}{2} \) and \( B = -\frac{1}{2} \). Thus, \( \frac{1}{y(y-2)} = \frac{1}{2} \left( \frac{1}{y} - \frac{1}{y-2} \right) \).
04

Integrate Both Sides

Integrate both sides: \( \int \left( \frac{1}{2} \left( \frac{1}{y} - \frac{1}{y-2} \right) \right) dy = \int dx \). This results in \( \frac{1}{2} (\ln |y| - \ln |y-2|) = x + C \).
05

Simplify and Solve for C

Simplify using logarithm properties: \( \ln \left| \frac{y}{y-2} \right| = 2x + C_1 \). Solve for \( C_1 \) using the initial condition \( y(0) = 1 \): \( \ln \left| \frac{1}{1-2} \right| = C_1 \), resulting in \( \ln |1| = C_1 = 0 \).
06

Solve for y

Equation becomes \( \ln \left| \frac{y}{y-2} \right| = 2x \). Exponentiating both sides gives \( \frac{y}{y-2} = e^{2x} \). Simplify to find \( y = \frac{2e^{2x}}{1-e^{2x}} \).

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Key Concepts

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

Separable Differential Equations
A separable differential equation is a type of differential equation that can be rearranged so that each variable appears on a different side of the equation. In the given problem, we have the equation \( \frac{d y}{d x} = y(y-2) \). This can be described as separable because the expression on the right-hand side depends solely on \( y \). To solve such equations, you need to:
  • Separate the variables by getting all terms involving \( y \) on one side and all terms involving \( x \) (including \( dx \)) on the other side of the equation.
  • This often involves algebraic manipulation like factoring out terms or dividing the entire expression by another term to isolate \( dy \) and \( dy \).
For our equation, it's done by rewriting it as \( \frac{d y}{y(y-2)} = dx \). After separating, each side can be integrated to eventually find the solution to the differential equation.
Initial Conditions
Initial conditions in differential equations provide the necessary values to find a specific solution to the equation. Here, the initial condition is given as \( y(0) = 1 \). This means when \( x = 0 \), \( y \) is equal to 1. When you solve a differential equation and reach the point where a constant of integration (often denoted as \( C \)) is present, initial conditions help you solve for this constant:
  • Substitute the initial condition values into the integrated equation to find the specific value of \( C \).
  • This ensures the solution satisfies the particular condition given, making the solution valid not just generally but specifically for the context discussed.
In the given problem, using the condition \( y(0) = 1 \), we find that the constant \( C \) resolves to zero, fine-tuning the solution to fit the context provided.
Integrating Differential Equations
Integration is a crucial step in solving differential equations, especially after separating the variables. During this process, we need to evaluate integrals such as \( \int \left( \frac{1}{2} \left( \frac{1}{y} - \frac{1}{y-2} \right) \right) dy = \int dx \).Here's how it generally works:
  • Apply integration to each separated term. For terms involving \( y \), integrate with respect to \( y \) and for terms with \( x \), integrate with respect to \( x \).
  • In this specific problem, we use partial fractions to split the term \( \frac{1}{y(y-2)} \) into two simpler fractions that are easier to integrate: \( \frac{1}{2}\int \left( \frac{1}{y} - \frac{1}{y-2} \right) dy \).
  • The integration results in terms involving natural logs, i.e., \( \frac{1}{2} (\ln |y| - \ln |y-2|) = x + C \).
After integration, handle the algebraic expressions to find the explicit formula for \( y \) in terms of \( x \), as illustrated with the final solution \( y = \frac{2e^{2x}}{1-e^{2x}} \). This makes integration essential in transitioning from a differential to an algebraic expression that represents the solution.

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Most popular questions from this chapter

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). \(\frac{d N}{d t}=N(N-1)(5-N)\) (a) \(N(0)=1\), (b) \(N(0)=1 / 2\), (c) \(N(0)=3 / 2\), (d) \(N(0)=7\).

Homeostasis Sterner and Elser (2002) studied the relationship between the amount of nitrogen in an animal's body and the amount of nitrogen present in the food that it eats. Many animals maintain homeostasis (balance), that is, they control their own nitrogen content. As the amount of nitrogen present in their food increases, the amount of nitrogen in the animal's body increases more slowly. If the amount of nitrogen in the animal is \(N\) and the amount of nitrogen in its food is \(F\), Sterner and Elser argue that: $$ \frac{1}{N} \frac{d N}{d t}=\frac{\sigma}{F} \frac{d F}{d t} $$ where \(\sigma\) is a constant. (a) Show that if \(\sigma=1\), then \(N \propto F ;\) that is, the nitrogen content of the animal increases in proportion to its food. This is called absence of homeostasis. (b) If \(\sigma=0\), then \(N\) is a constant, independent of \(F\). This is called homeostasis (the animal maintains a balanced amount of nitrogen, independent of its food). (c) Show that if \(0<\sigma<1\), then, if \(F\) doubles, \(N\) also increases but by a factor less than 2 .

You should treat \(h\) as a constant. For what values of \(h\) (if any) does each equation have equilibria? Use a graphical argument to show which of the equilibria (if any) are stable. $$ \frac{d y}{d x}=(y-1)(y+3)-h $$

For Problems \(49-56\) determine whether the equilibrium at \(x=0\) is stable, unstable, or semi-stable. $$ \frac{d x}{d t}=x^{3} $$

One of the key ideas for sketching solutions from vector field plots is that a solution curve must be monotonic, that is, \(x(t)\) is either increasing or decreasing or constant but cannot switch from one behavior to another. We showed that a solution \(x(t)\) could not start by increasing and then switch to decreasing. Suppose that \(x(t)\) is a solution of the differential equation \(\frac{d x}{d t}=g(x)\) and that \(x(t)\) starts off decreasing with time. Show that \(x(t)\) cannot switch to increasing.
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