/*! 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 29 Consider the equation $$\left(y... [FREE SOLUTION] | 91Ó°ÊÓ

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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?

Short Answer

Expert verified
After initial examination, the given differential equation was not exact. However, upon multiplying by the integrating factor \( y^{-2} \), it became exact and could be solved using standard exact differential equation procedures. Unfortunately, applying the integrating factor lost solutions corresponding to \( y = 0 \).

Step by step solution

01

Check for Exactness

Take the given equation, \( (y^{2} + 2xy)dx - x^{2}dy = 0 \). This is in the form of \( Mdx + Ndy = 0 \). To check for exactness, we calculate \( \frac{dM}{dy} \) and \( \frac{dN}{dx} \) and see if they are equal. \( M = y^{2} + 2xy \) and \( N = -x^{2} \). Thus, \( \frac{dM}{dy} = 2y + 2x \) and \( \frac{dN}{dx} = -2x \). Since \( \frac{dM}{dy} \) is not equal to \( \frac{dN}{dx} \), the given differential equation is not exact.
02

Find the Integrating Factor

It is given that multiplying both sides by \( y^{-2} \) makes the equation exact. So, multiply the equation \( (y^{2} + 2xy)dx - x^{2}dy = 0 \) by \( y^{-2} \) to get \( (1 + 2x/y)dx - (x^{2}/y^{2})dy = 0 \), where the new \( M = 1 + 2x/y \) and \( N = -x^{2}/y^{2} \). Now, calculate \( \frac{dM}{dy} \) and \( \frac{dN}{dx} \). You will find that they are equal, thus the new equation is exact.
03

Solve the Exact Differential Equation

Now, with the exact equation, \( (1 + 2x/y)dx - (x^{2}/y^{2})dy = 0 \), we can form \( \phi(x, y) = constant \), where \( \phi(x, y) \) is the solution to the exact differential equation obtained by integrating \( M \) with respect to \( x \) and \( N \) with respect to \( y \).
04

Analyze Lost Solutions

When an integrating factor was applied, all the solutions corresponding to \( y = 0 \) were lost, as it involved dividing by \( y \). Therefore, the solution to the original non-exact differential equation must include a flat solution at \( y = 0 \).

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Key Concepts

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

Integrating Factor
When working with differential equations, sometimes the equation might not be exact, making it challenging to solve directly. This is where the integrating factor comes into play. An integrating factor is like a magic multiplier that transforms a non-exact equation into an exact one, where solving becomes straightforward.

To find this factor, you typically need to identify a function or term that, when multiplied with the entire equation, aligns the partial derivatives in a way that satisfies the exactness condition. In our exercise, multiplying both sides of the equation by \( y^{-2} \) changed the original non-exact equation into an exact one. This transformation is crucial because it simplifies the process of finding the solution by making the mathematics more manageable.

Keep in mind:
  • Identifying the integrating factor can vary in complexity.
  • It involves making partial derivatives equal.
  • Once exact, solving becomes simply a matter of integration along specific paths.
Non-exact Differential Equation
Not all differential equations are born equal; sometimes they are non-exact, and this presents a hurdle. A non-exact differential equation is one where the given form \( Mdx + Ndy = 0 \) does not satisfy the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). In simpler terms, the equation is not perfectly balanced in terms of the mixed derivatives.

For instance, in our exercise, the initial equation was non-exact because the partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) did not match. This mismatch indicates asymmetry that needs to be adjusted, often with the use of an integrating factor or through other methods, to transform the equation into an exact form. Understanding why an equation is non-exact is the first step in adjusting and solving it correctly.

Key points about non-exact differential equations:
  • They do not directly fulfill the exactness condition.
  • A solution path involves making the equation exact first.
  • Partial derivatives \( \frac{dM}{dy} \) and \( \frac{dN}{dx} \) do not align.
Exactness Condition
The exactness condition in differential equations is a crucial check that determines the path to obtaining a solution. This condition states that the differential equation in the standard form \( Mdx + Ndy = 0 \) is considered exact if the partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) are equal. When this condition is met, it means that there exists a potential function \( \phi(x, y) \), such that the derivative \( abla \phi = (M, N) \).

For instance, in our solution, by employing the integrating factor \( y^{-2} \), we ensured equality between \( \frac{dM}{dy} \) and \( \frac{dN}{dx} \), thereby satisfying the condition for exactness. This equivalence is what ultimately allows us to solve the differential equation by integration without additional complications.

Why is exactness crucial?
  • Exactness guarantees a potential function solution.
  • It allows the use of straightforward methods like integration.
  • Ensures the symmetry needed for solving.
Lost Solutions in Integration
While focusing on transforming and solving a differential equation, it is easy to overlook what is left behind. 'Lost solutions' refer to the solutions that do not appear in the final answer due to restrictions in the process, like division by zero. This often happens when using integrating factors.

In the exercise, applying the integrating factor \( y^{-2} \) inadvertently excluded solutions where \( y = 0 \) because division by zero is not allowed. The transformation simplifies unknowns but at a potential cost: some real, but specific solutions are immeasurable in the modified process.

To handle lost solutions, it is essential to:
  • Identify them post-solution by revisiting the original differential equation.
  • Account for all possible solutions, including those constrained by initial conditions.
  • Seek generality in the final solution to cover lost cases.
Understanding and addressing lost solutions ensures a comprehensive solution that aligns with all initial conditions and constraints posed in the original problem.

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Most popular questions from this chapter

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 \( \pm 2 \).

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.]

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.

In Problems \(1-6,\) determine whether the given differential equation is separable. $$ \frac{d y}{d x}=4 y^{2}-3 y+1 $$

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.)

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