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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 \(n=1\).

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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.

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Most popular questions from this chapter

Mixing. Suppose a brine containing 0.3 kilogram (kg) of salt per liter (L) runs into a tank initially filled with 400 L of water containing 2 kg of salt. If the brine enters at 10 L/min, the mixture is kept uniform by stirring, and the mixture flows out at the same rate. Find the mass of salt in the tank after 10 min (see Figure 2.4). [Hint: Let A denote the number of kilograms of salt in the tank at t min after the process begins and use the fact that rate of increase in $$A =$$ rate of input - rate of exit. A further discussion of mixing problems is given in Section 3.2.]

Free Fall. In Section 2.1, we discussed a model for an object falling toward Earth. Assuming that only air resistance and gravity are acting on the object, we found that the velocity \( \boldsymbol{v} \) must satisfy the equation $$ m \frac{d v}{d t}=m g-b v $$ where \( m \) is the mass, \( \boldsymbol{g} \) is the acceleration due to gravity, and \( b>0 \) is a constant (see Figure 2.1). If \( m=100 \mathrm{kg} \), \( g=9.8 \mathrm{m} / \mathrm{sec}^{2}, b=5 \mathrm{kg} / \mathrm{sec} \) and \( v(0)=10 \mathrm{m} / \mathrm{sec} \), solve for \( v(t) \). What is the limiting (i.e., terminal) velocity of the object?

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.)

Interval of Definition. By looking at an initial value problem \( d y / d x=f(x, y) \) with \( y\left(x_{0}\right)=y_{0} \), it is not always possible to determine the domain of the solution \( y(x) \) or the interval over which the function \( y(x) \) satisfies the differential equation. (a) Solve the equation \( d y / d x=x y^{3} \). (b) Give explicitly the solutions to the initial value problem with \( y(0)=1 ; y(0)=1 / 2 ; y(0)=2 \). (c) Determine the domains of the solutions in part (b). (d) As found in part (c), the domains of the solutions depend on the initial conditions. For the initial value problem \( d y / d x=x y^{3} \) with \( y(0)=a, a>0 \), show that as a approaches zero from the right the domain approaches the whole real line \( (-\infty, \infty) \) and as \( a \) approaches \( +\infty \) the domain shrinks to a single point. (e) Sketch the solutions to the initial value problem \( d y / d x=x y^{3} \) with \( y(0)=a \) for \( a=\pm 1 / 2, \pm 1 \), and \( \pm 2 \).

$$\frac{d y}{d x}=\frac{y(\ln y-\ln x+1)}{x}$$
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