/*! 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 42 Solving initial value problems S... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solving initial value problems Solve the following initial value problems. $$p^{\prime}(x)=\frac{2}{x^{2}+x}, p(1)=0$$

Short Answer

Expert verified
Question: Determine the particular solution to the initial value problem \(p'(x) = \frac{2}{x^2 + x}\) with \(p(1) = 0\). Answer: The particular solution to the given initial value problem is: \(p(x) = 2\ln|x| - 2\ln|x+1| + 2\ln|2|\).

Step by step solution

01

Integrate both sides of the differential equation

Integrate both sides of the given differential equation: $$\int p'(x) \, dx = \int \frac{2}{x^2 + x} \, dx.$$
02

Separate the right-side fraction into partial fractions

To integrate the right-side fraction, we first need to rewrite it as the sum of partial fractions. Since the denominator is \((x)(x+1)\), we can rewrite the fraction with the form: $$\frac{2}{x^2 + x} = \frac{A}{x} + \frac{B}{x+1}.$$ Now we can set up a system of equations to solve for A and B.
03

Solve for A and B using cross-multiplication

Multiply both sides of the equation by \(x^2 + x\) to cancel out the denominators: $$2 = A(x+1) + B(x).$$ Now set up the system of equations to find A and B by letting \(x=0\) and \(x=-1\). When \(x=0\), we get the equation \(2 = A\), so \(A=2\). When \(x=-1\), we get the equation \(2 = -B\), so \(B=-2\). Thus, we have the partial fractions as $$\frac{2}{x^2 + x} = \frac{2}{x} - \frac{2}{x+1}.$$
04

Integrate the partial fractions

Now integrate the right-side partial fractions: $$\int p'(x) \, dx = \int \left(\frac{2}{x} - \frac{2}{x+1}\right) \, dx = 2\int\frac{1}{x} \, dx - 2\int\frac{1}{x+1} \, dx.$$ Apply the integral of the natural logarithm: $$p(x) = 2\ln|x| - 2\ln|x+1| + C,$$ where C is the integration constant.
05

Use the initial condition to find the particular solution

Now, we will use the initial condition \(p(1)=0\) to find the value of C: $$0 = 2\ln|1| - 2\ln|1+1| + C.$$ Since \(\ln|1|=0\) and \(\ln|2|\neq 0\), we get: $$0 = -2\ln|2| + C \Rightarrow C = 2\ln|2|.$$
06

Write the final particular solution

Combine the general solution and the value of C to get the particular solution for the initial value problem: $$p(x) = 2\ln|x| - 2\ln|x+1| + 2\ln|2|.$$ This is the final solution to the given initial value problem.

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 91Ó°ÊÓ!

Key Concepts

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

Differential Equations
Differential equations are like equations in math that involve functions and their derivatives. Essentially, they show the relationship between a function and the rates at which it changes. These equations come in handy when trying to model complex systems in fields like physics, engineering, and economics. There are different types of differential equations, such as ordinary (ODEs) and partial (PDEs), based on the number of independent variables they involve. In our case, the differential equation we are working with is an ordinary differential equation because it involves a single independent variable, \( x \). By solving a differential equation, we seek a function that satisfies the equation's constraints, which often include an initial condition too. This initial condition, provided at a specific point, allows us to find a unique solution that otherwise would include an unknown constant.
Partial Fractions
Partial fractions make life easier when we need to integrate complex rational expressions, especially those with polynomial denominators. Instead of dealing directly with a complicated fraction, we break it down into simpler parts called partial fractions. This process, known as decomposition, expresses a complex fraction as the sum of simpler fractions.
  • The target is to have terms like \( \frac{A}{x} \) and \( \frac{B}{x+1} \), which are easier to integrate.
  • We begin by identifying the factors of the polynomial in the denominator.
  • Next, we express the given fraction as a sum based on these factors.
  • Finally, we solve for the constants \( A \) and \( B \) using algebraic methods.
For instance, in the given problem, the expression \( \frac{2}{x^2 + x} \) was decomposed into \( \frac{2}{x} - \frac{2}{x+1} \). This made the integration process straightforward.
Integration
Integration is the reverse process of differentiation. It allows us to find the function from its derivative. When dealing with integrals, especially indefinite ones, we're always searching for a family of functions that can represent the anti-derivative.
  • In the context of solving differential equations, integration helps us move from a derivative back to the original function.
  • For rational expressions like those obtained from partial fraction decomposition, integration often involves natural logarithms.
In the example problem, integrating \( \frac{2}{x} - \frac{2}{x+1} \) leads to the expression \( 2\ln|x| - 2\ln|x+1| + C \). Here, \( C \) is an arbitrary constant, indicating an entire set of functions that can be solutions. It's this process that restores the function from its rate of change.
Initial Condition
Initial conditions are critical in determining a specific solution from the endless possibilities present in the general solution of a differential equation. When integrating, we introduce an arbitrary constant, \( C \). This constant is adjusted based on the initial condition to find the particular solution relevant to the problem at hand.
  • An initial value problem provides an extra piece of information, like \( p(1) = 0 \), which fixes \( C \).
  • This condition ensures the solution curve passes through a specified point.
  • By plugging values into the general solution, we can solve for \( C \) and determine the exact form of our function.
Using the initial condition \( p(1) = 0 \), we computed \( C \) to be \( 2\ln|2| \). This adjusted our solution to \( p(x) = 2\ln|x| - 2\ln|x+1| + 2\ln|2| \), completing our task of identifying the exact behavior of the system described by the differential equation.

One App. One Place for Learning.

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

Get started for free

Most popular questions from this chapter

A special class of first-order linear equations have the form \(a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=f(t),\) where \(a\) and \(f\) are given functions of t. Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form $$a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=\frac{d}{d t}(a(t) y(t))=f(t)$$ Therefore, the equation can be solved by integrating both sides with respect to \(t .\) Use this idea to solve the following initial value problems. $$e^{-t} y^{\prime}(t)-e^{-t} y=e^{2 t}, y(0)=4$$

Let \(y(t)\) be the population of a species that is being harvested, for \(t \geq 0 .\) Consider the harvesting model \(y^{\prime}(t)=0.008 y-h, y(0)=y_{0},\) where \(h\) is the annual harvesting rate, \(y_{0}\) is the initial population of the species, and \(t\) is measured in years. a. If \(y_{0}=2000,\) what harvesting rate should be used to maintain a constant population of \(y=2000,\) for \(t \geq 0 ?\) b. If the harvesting rate is \(h=200 /\) year, what initial population ensures a constant population?

U.S. population projections According to the U.S. Census Bureau, the nation's population (to the nearest million) was 296 million in 2005 and 321 million in \(2015 .\) The Bureau also projects a 2050 population of 398 million. To construct a logistic model, both the growth rate and the carrying capacity must be estimated. There are several ways to estimate these parameters. Here is one approach: a. Assume \(t=0\) corresponds to 2005 and that the population growth is exponential for the first ten years; that is, between 2005 and \(2015,\) the population is given by \(P(t)=P(0) e^{\pi}\) Estimate the growth rate \(r\) using this assumption. b. Write the solution of the logistic equation with the value of \(r\) found in part (a). Use the projected value \(P(45)=398\) million to find a value of the carrying capacity \(K\) c. According to the logistic model determined in parts (a) and (b), when will the U.S. population reach \(95 \%\) of its carrying capacity? d. Estimations of this kind must be made and interpreted carefully. Suppose the projected population for 2050 is 410 million rather than 398 million. What is the value of the carrying capacity in this case? e. Repeat part (d) assuming the projected population for 2050 is 380 million rather than 398 million. What is the value of the carrying capacity in this case? E. Comment on the sensitivity of the carrying capacity to the 35-year population projection.

The following initial value problems model the payoff of a loan. In each case, solve the initial value problem, for \(t \geq 0,\) graph the solution, and determine the first month in which the loan balance is zero. $$B^{\prime}(t)=0.01 B-750, B(0)=45,000$$

Chemical rate equations Let \(y(t)\) be the concentration of a substance in a chemical reaction (typical units are moles/liter). The change in the concentration, under appropriate conditions, is modeled by the equation \(\frac{d y}{d t}=-k y^{n},\) for \(t \geq 0,\) where \(k>0\) is a rate constant and the positive integer \(n\) is the order of the reaction. a. Show that for a first-order reaction \((n=1),\) the concentration obeys an exponential decay law. b. Solve the initial value problem for a second-order reaction \((n=2)\) assuming \(y(0)=y_{0}\). c. Graph the concentration for a first-order and second-order reaction with \(k=0.1\) and \(y_{0}=1\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.