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True or False: Every separable first-order equation \(d y / d x=g(x) h(y)\) is exact.

Short Answer

Expert verified
False, not every separable first-order equation is exact.

Step by step solution

01

Understanding the Problem

The exercise asks us whether every separable first-order differential equation is also an exact equation. A separable equation is of the form \( \frac{dy}{dx} = g(x)h(y) \), and an exact equation is one that can be expressed as \( M(x, y)dx + N(x, y)dy = 0 \) with a specific condition on the partial derivatives \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
02

Reviewing Separable Equations

A separable differential equation can be rearranged to \( \frac{1}{h(y)}dy = g(x)dx \) and then integrated on both sides. The key characteristic is that the expression on each side depends only on one variable (\(x\) or \(y\)).
03

Reviewing Exact Equations

An exact equation satisfies the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). This means there should be a function \( \Psi(x, y) \) such that its total differential \( d\Psi = M dx + N dy \).
04

Verifying Exactness in Separable Equations

To determine if a separable equation is exact, consider it in the form \( h(y) dy + (-g(x) dx) = 0 \). Here, \( M = h(y) \) and \( N = -g(x) \). For exactness, \( \frac{\partial h(y)}{\partial x} = 0 \) and \( \frac{\partial -g(x)}{\partial y} = 0 \), which are true, but the cross partial derivatives needed to check exactness would force both to be constant zero, which does not match the original forms \( h(y) \) and \( g(x) \) unless trivially both \(h(y)\) and \(g(x)\) are constants.
05

Conclusion

With the conditions for exactness violated unless both functions are constants, we conclude that not every separable equation can be exact given arbitrary non-constant functions.

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Key Concepts

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

Separable Equations
Separable equations are a special type of first-order differential equations. These equations have the form \( \frac{dy}{dx} = g(x)h(y) \). The main feature of separable equations is that each side of the equation can be expressed in terms of different variables. One side will only contain terms involving \(x\) and the other side will only involve terms of \(y\). This separation of variables allows these equations to be easily integrated.

To solve a separable equation, rearrange it to \( \frac{1}{h(y)}dy = g(x)dx \). Then, integrate both sides separately. This results in solutions, often in the implicit form, like \( F(y) = G(x) + C \), where \(C\) is the integration constant. Whether you can solve the equation explicitly for \(y\) depends on the functions involved, but often these can be manipulated to find a particular solution.

Separable equations are prevalent in real-world scenarios, such as modeling exponential growth or decay, where a system's change depends on both an independent variable and the system's current state.
Exact Equations
Exact equations are another kind of first-order differential equations. They have the general form \( M(x, y)dx + N(x, y)dy = 0 \). The defining feature of an exact equation is that there exists a function \( \Psi(x, y) \) such that its total differential equals the expression \( d\Psi = Mdx + Ndy \). For a differential equation to be exact, a specific condition must be satisfied: the partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) must be equal.

In practical terms, if this condition is met, it means the original differential equation can be expressed as the derivative of some potential function \( \Psi(x, y) \). Solving the equation involves finding \( \Psi(x, y) \) such that \( \Psi(x, y) = C \), where \(C\) is a constant. Typically, you might integrate \(M\) with respect to \(x\) and \(N\) with respect to \(y\), then combine results, taking care to consider the cross partial derivatives.

Exactness in equations is crucial because it provides a straightforward way to solve potentially complex equations by recognizing they can be boiled down to finding and differentiating a single function.
First-Order Differential Equation
A first-order differential equation is one where the highest derivative present is a first derivative. These equations take various forms and model numerous physical, biological, and engineering systems. In mathematical notation, they generally look like \( \frac{dy}{dx} = f(x, y) \). They describe the rate of change of a variable \(y\) in relation to \(x\).

First-order equations can be classified into multiple subtypes, including separable equations and exact equations. Each subtype offers a distinct method of solution depending on specific characteristics of the equation.
  • Separable Equations: These can be solved by separating variables, as mentioned earlier, and integrating both sides.
  • Exact Equations: These rely on the comparison of partial derivatives and often lead to potential functions.

Understanding the form and solving technique for each type of first-order differential equation is essential for effectively modeling and predicting the behavior of various systems.

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Most popular questions from this chapter

When forgetfulness is taken into account, the rate of memorization of a subject is given by $$ \frac{d A}{d t}=k_{1}(M-A)-k_{2} A $$ where \(k_{1}>0, k_{2}>0, A(t)\) is the amount to be memorized in time \(t, M\) is the total amount to be memorized, and \(M-A\) is the amount remaining to be memorized. See Problems 25 and 26 in Exercises \(1.3\). (a) Since the DE is autonomous, use the phase portrait concept of Section \(2.1\) to find the limiting value of \(A(t)\) as \(t \rightarrow \infty\). Interpret the result. (b) Solve for \(A(t)\) subject to \(A(0)=0\). Sketch the graph of \(A(t)\) and verify your prediction in part (a).

The population of a town grows at a rate proportional to the population present at time \(t\). The initial population of 500 increases by \(15 \%\) in 10 years. What will the population be in 30 years? How fast is the population growing at \(t=30 ?\)

(a) The Fresnel sine integral is defined by \(S(x)=\) \(\int_{0}^{x} \sin \left(\pi t^{2} / 2\right) d t\). Express the solution \(y(x)\) of the initialvalue problem \(y^{\prime}-\left(\sin x^{2}\right) y=0, y(0)=5\), in terms of \(S(x)\) (b) Use a CAS to graph the solution curve for the IVP on \((-\infty, \infty)\) (c) It is known that \(S(x) \rightarrow \frac{1}{2}\) as \(x \rightarrow \infty\) and \(S(x) \rightarrow-\frac{1}{2}\) as \(x \rightarrow-\infty\). What does the solution \(y(x)\) approach as \(x \rightarrow \infty ?\) As \(x \rightarrow-\infty ?\) (d) Use a CAS to find the values of the absolute maximum and the absolute minimum of the solution \(y(x)\).

Solve the given initial-value problem by finding, as in Example 4, an appropriate integrating factor. $$ \left(x^{2}+y^{2}-5\right) d x=(y+x y) d y, \quad y(0)=1 $$

A dead body was found within a closed room of a bouse where the temperature was a constant \(70^{\circ} \mathrm{F}\). At the time of discovery, the core temperature of the body was determined to be \(85^{\circ} \mathrm{F}\). One hour later a second measurement showed that the core temperature of the body was \(80^{\circ} \mathrm{F}\). Assume that the time of death corresponds to \(t=0\) and that the core temperature at that time was \(98.6^{\circ} \mathrm{F}\). Determine how many hours elapsed before the body was found.

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