/*! 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 48 Derive the following general for... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 48

Derive the following general formula for the solution to the Bernoulli equation (9): $$y = \left\\{ \begin{array} { l l } { \left[ \frac { ( 1 - n ) \int e ^ { ( 1 - n ) \int P ( x ) d x } Q ( x ) d x + C _ { 1 } ] ^ { 1 / ( 1 - n ) } } { e ^ { ( 1 - n ) \int P ( x ) d x } Q ( x ) d x + C _ { 1 } } \right] ^ { 1 / ( 1 - n ) } } & { \text { for } n \neq 1 } \\ { C _ { 2 } e ^ { \int [ Q ( x ) - P ( x ) ] d x } } & { \text { for } n = 1 } \end{array} \right.$$

Short Answer

Expert verified
The general solution to a Bernoulli equation is: \(y = \left[ e^{-\int P(x) dx} \left[ C1 + \int e^{\int P(x) dx} Q(x) dx \right] \right]^{1/(1-n)}\) for \(n \neq 1\), and \(y = C2 e^{\int [Q(x) - P(x)] dx}\) for \(n = 1\).

Step by step solution

01

Derive the solution for \(n\neq1\)

Start by transforming the Bernoulli equation to a linear differential equation through the following substitution: \(v=y^{1-n}\). Then apply the following operator to both sides of the Bernoulli equation: \((1-n)v' = Q(x) - P(x)v\). This is now a linear first-order homogeneous differential equation, the solution of which is \(v = e^{-\int P(x) dx} \left[ C1 + \int e^{\int P(x) dx} Q(x) dx \right]\). Reverse the substitution to find \(y = \left[ e^{-\int P(x) dx} \left[ C1 + \int e^{\int P(x) dx} Q(x) dx \right] \right]^{1/(1-n)}\).
02

Derive the solution for \(n=1\)

When \(n=1\), the Bernoulli equation simplifies to a linear first-order homogeneous differential equation: \(y' + P(x)y = Q(x)\). The solution to this can be written directly using an integrating factor, yielding \(y = C2 e^{\int [Q(x) - P(x)] dx}\).
03

Present the general solution of a Bernoulli equation

The general solution for a Bernoulli equation is then \(y\) equals to the solution derived in Step 1 for \(n\neq1\) and the solution derived in Step 2 for \(n=1\).

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.

Linear Differential Equations
Linear differential equations are a fundamental concept in mathematics, often encountered in problems that involve rates of change. They have the standard form \( y' + P(x)y = Q(x) \), where \( y' \) is the derivative of \( y \) with respect to \( x \). The terms \( P(x) \) and \( Q(x) \) represent known functions of \( x \), which makes these equations linear because the function \( y \) and its derivatives appear to the first power only.
A special feature of linear differential equations is the possibility of being either inhomogeneous or homogeneous.
  • An equation is homogeneous if \( Q(x) = 0 \).
  • Otherwise, when \( Q(x) eq 0 \), it is inhomogeneous.
A key reason for solving linear differential equations is that they have well-established methods for obtaining solutions, one of which includes the use of an integrating factor that simplifies the solution process.
Homogeneous Differential Equations
Homogeneous differential equations are a specific category where the right-hand side function belongs to a class that incorporates consistency in mathematical structure. They have the form \( y' + P(x)y = 0 \).
These equations have unique characteristics:
  • The entire equation can be reduced by dividing through by any non-zero variable, maintaining its structure.
  • They often result in solutions that describe exponential growth or decay.
The significance of homogeneous equations lies in their simplicity because solutions can generally be expressed as a product of an exponential function by solving them as if they are separable differential equations.
Integrating Factor
The integrating factor is an essential tool used for solving linear differential equations. It simplifies the equation to make it more manageable.
Here's how it works:
  • The integrating factor is usually denoted as \( \, \mu(x) = e^{\int P(x) dx} \).
  • Multiplying every term in the differential equation by this integrating factor transforms the left-hand side into the derivative of a product.
For a given linear equation \( y' + P(x)y = Q(x) \), multiplying by the integrating factor yields:
\( \frac{d}{dx}[\mu(x)y] = \mu(x)Q(x) \).
On integrating both sides with respect to \( x \), we eventually solve for the function \( y(x) \). This method streamlines solving otherwise challenging linear differential equations into a straightforward process.
Differential Equation Solutions
Solutions to differential equations often involve determining a function that satisfies the equation for all values within a certain interval. When dealing with specific types of equations, like the Bernoulli equation, solutions may require special approaches.
In the case of the Bernoulli equation, solutions differentiate based on whether \( n \) equals 1 or not.
  • For \( n eq 1 \), substitution transforms the equation into one solvable using linear differential equation methods.
  • For \( n = 1 \), it becomes a more typical linear differential equation handled by the integrating factor technique.
The elegance of differential equation solutions is in absorbing complex expressions through methodical, smart substitutions and manipulations, culminating in a clear expression for the variable of interest, offering insights into natural phenomena and processes.

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

In Problems \(1-8,\) classify the equation as separable, linear, exact, or none of these. Notice that some equations may have more than one classification. $$x y d x+d y=0$$

Consider the equation $$\left(y^{2}+2 x y\right) d x-x^{2} d y=0$$ (a) Show that this equation is not exact. (b) Show that multiplying both sides of the equation by $$y^{-2}$$ yields a new equation that is exact. (c) Use the solution of the resulting exact equation to solve the original equation. (d) Were any solutions lost in the process?

Compound Interest. If \(P(t)\) is the amount of dollars in a savings bank account that pays a yearly interest rate of \(r \%\) compounded continuously, then $$\frac{d P}{d t}=\frac{r}{100} P, \quad t$$ in years. Assume the interest is 5\(\%\) annually, \(P(0)=\$ 1000,\) and no monies are withdrawn. (a) How much will be in the account after 2 yr? (b) When will the account account every 12 months, (c) If \(\$ 1000\) is added to the account every 12 months, how much will be in the account after 3\(\frac{1}{2}\) yr?

Uniqueness Questions. In Chapter 1 we indicated that in applications most initial value problems will have a unique solution. In fact, the existence of unique solutions was so important that we stated an existence and uniqueness theorem, Theorem 1, page 11. The method for separable equations can give us a solution, but it may not give us all the solutions (also see Problem 30). To illustrate this, consider the equation \( d y / d x=y^{1 / 3} \). (a) Use the method of separation of variables to show that $$ y=\left(\frac{2 x}{3}+C\right)^{3 / 2} $$ is a solution. (b) Show that the initial value problem \( d y / d x \) = \( y^{1 / 3} \) with \( y(0)=0 \) is satisfied for \( C=0 \) by \( y=(2 x / 3)^{3 / 2} \) for \( x \geq 0 \). (c) Now show that the constant function \( y \equiv 0 \) also satisfies the initial value problem given in part (b). Hence, this initial value problem does not have a unique solution. (d) Finally, show that the conditions of Theorem 1 on page 11 are not satisfied. (The solution \( y \equiv 0 \) was lost because of the division by zero in the separation process.)

As discussed in calculus, certain indefinite integrals (antiderivatives) such as \( \int e^{x^{x}} d x \) cannot be expressed in finite terms using elementary functions. When such an integral is encountered while solving a differential equation, it is often helpful to use definite integration (integrals with variable upper limit). For example, consider the initial value problem $$ \frac{d y}{d x}=e^{x^{2}} y^{2}, \quad y(2)=1 $$ The differential equation separates if we divide by \( y^{2} \) and multiply by \( d x \). We integrate the separated equation from \( x=2 \) to \( x=x_{1} \) and find $$ \int_{x=2}^{x=x_{1}} e^{x^{2}} d x=\int_{x=2}^{x=x_{1}} \frac{d y}{y^{2}} $$ $$ =-\frac{1}{y} \left| \begin{array}{l}{x=x_{1}} \\ {x=2}\end{array}\right. $$ $$ =-\frac{1}{y\left(x_{1}\right)}+\frac{1}{y(2)} $$ If we let \( t \) be the variable of integration and replace \( x_{1} \) by \( x \) and \( y(2) \) by 1, then we can express the solution to the initial value problem by $$ y(x)=\left(1-\int_{2}^{x} e^{t^{2}} d t\right)^{-1} $$ Use definite integration to find an explicit solution to the initial value problems in parts (a)-(c). (a) \( d y / d x=e^{x^{2}}, \quad y(0)=0 \) (b) \( d y / d x=e^{x^{2}} y^{-2}, \quad y(0)=1 \) (c) \( d y / d x=\sqrt{1+\sin x}\left(1+y^{2}\right), \quad y(0)=1 \) (d) Use a numerical integration algorithm (such as Simpson's rule, described in Appendix C) to approximate the solution to part (b) at \( x=0.5 \) to three decimal places.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation
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.