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Determine whether or not each of the equations is exact. If it is exact, find the solution. $$ (2 x+4 y)+(2 x-2 y) y^{\prime}=0 $$

Short Answer

Expert verified
If so, what is the solution? Answer: The given equation is not exact, as the partial derivatives of M with respect to y and N with respect to x are not equal (\(\frac{\partial M}{\partial y} = 4\) and \(\frac{\partial N}{\partial x} = 2\)). Therefore, we cannot find a solution using the exact differential equation method.

Step by step solution

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01

Identify M(x, y) and N(x, y)

We need to identify the terms M(x, y) and N(x, y) from the given equation. Our equation is \((2x + 4y) + (2x - 2y)y' = 0\), so: $$M(x, y) = 2x + 4y$$ $$N(x, y) = 2x - 2y$$
02

Check if the equation is exact

Let's check if the equation is exact by computing the partial derivatives of M with respect to y and of N with respect to x: $$\frac{\partial M}{\partial y} = 4$$ $$\frac{\partial N}{\partial x} = 2$$ Since \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}\), the given equation is not exact. Since the equation is not exact, we cannot proceed further to find the solution as per the method mentioned in the analysis. There might be other methods to solve this equation, or it may not have a general solution.

Key Concepts

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

Ordinary Differential Equations
Ordinary differential equations (ODEs) form the backbone of many mathematical, scientific, and engineering inquiries. At their core, ODEs involve functions and their derivatives, exposing the relationships between quantities and their rates of change.

An ODE can often be recognized by its structure: it will involve an unknown function, often denoted as y(x) or y(t), where the independent variable x or t represents elements such as space or time, and its derivatives, symbolized as y', y'', etc. When we look at an ODE, we're looking to find a function that satisfies the equation for all values of x or t within a certain interval.

For instance, the linear first-order ODE such as y' + p(x)y = q(x), outlines how y changes with x when influenced by p(x) and q(x). The ODE provided in our exercise, involving both x, y, and the derivative of y with respect to x, denotes a kind of ODE known as a first-order differential equation because it involves only the first derivative of the function.
Method of Solution for ODEs
There are various methods to solve Ordinary Differential Equations (ODEs), depending on their form and complexity. For first-order ODEs, we often seek an analytical solution that involves separation of variables, integrating factors, or, as in the exercise, testing for exactness.

One particularly elegant method is to solve exact differential equations. An equation is called 'exact' if it can be expressed in the form of a total differential of a function \Phi(x, y), such that \(d\Phi = M(x, y)dx + N(x, y)dy = 0\). To check for exactness, we compare the partial derivative of M with respect to y to the partial derivative of N with respect to x. If these partial derivatives are equal, the equation is exact and there exists a potential function \Phi from which we can retrieve the solution to the ODE by integrating M and N appropriately.

In scenarios where an ODE is not exact, we may attempt to make it exact by multiplying an integrating factor, or we could use alternative methods such as power series expansion, Laplace transforms, or numerical approximation techniques for those that resist analytical methods.
Partial Derivatives
Partial derivatives are at the heart of multivariable calculus, and they play an instrumental role in the study of differential equations. When we are dealing with functions of more than one variable, such as \Phi(x, y), the rate at which \Phi changes with respect to one variable while holding others constant is given by its partial derivative.

In the context of exact differential equations, the role of partial derivatives is critical. As demonstrated in the exercise, the determination of whether an ODE is exact hinges on the equality of partial derivatives \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\). These derivatives indicate how each component of the equation changes in response to one variable at a time while keeping the other fixed.

The interesting thing about partial derivatives is that, while they reveal local behavior (how a function changes instantaneously with respect to one variable), they can also hint at a more global structure, such as potential functions or conservation laws, enabling us to solve complex ODEs.

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Most popular questions from this chapter

Show that if \(\left(N_{x}-M_{y}\right) / M=Q,\) where \(Q\) is a function of \(y\) only, then the differential equation $$ M+N y^{\prime}=0 $$ has an integrating factor of the form $$ \mu(y)=\exp \int Q(y) d y $$

let \(\phi_{0}(t)=0\) and use the method of successive approximations to approximate the solution of the given initial value problem. (a) Calculate \(\phi_{1}(t), \ldots, \phi_{4}(t),\) or (if necessary) Taylor approximations to these iterates. Keep tems up to order six. (b) Plot the functions you found in part (a) and observe whether they appear to be converging. Let \(\phi_{n}(x)=x^{n}\) for \(0 \leq x \leq 1\) and show that $$ \lim _{n \rightarrow \infty} \phi_{n}(x)=\left\\{\begin{array}{ll}{0,} & {0 \leq x<1} \\ {1,} & {x=1}\end{array}\right. $$

Harvesting a Renewable Resource. Suppose that the population \(y\) of a certain species of fish (for example, tuna or halibut) in a given area of the ocean is described by the logistic equation $$ d y / d t=r(1-y / K) y . $$ While it is desirable to utilize this source of food, it is intuitively clear that if too many fish are caught, then the fish population may be reduced below a useful level, and possibly even driven to extinction. Problems 20 and 21 explore some of the questions involved in formulating a rational strategy for managing the fishery. In this problem we assume that fish are caught at a constant rate \(h\) independent of the size of the fish population. Then \(y\) satisfies $$ d y / d t=r(1-y / K) y-h $$ The assumption of a constant catch rate \(h\) may be reasonable when \(y\) is large, but becomes less so when \(y\) is small. (a) If \(hy_{0}>y_{1},\) then \(y \rightarrow y_{2}\) as \(t \rightarrow \infty,\) but that if \(y_{0}r K / 4,\) show that \(y\) decreases to zero as \(l\) increases regardless of the value of \(y_{0}\). (c) If \(h=r K / 4\), show that there is a single cquilibrium point \(y=K / 2\) and that this point is semistable (see Problem 7 ). Thus the maximum sustainable yield is \(h_{m}=r K / 4\) corresponding to the equilibrium value \(y=K / 2 .\) Observe that \(h_{m}\) has the same value as \(Y_{m}\) in Problem \(20(\mathrm{d})\). The fishery is considered to be overexploited if \(y\) is reduced to a level below \(K / 2\).

(a) Solve the Gompertz equation $$ d y / d t=r y \ln (K / y) $$ subject to the initial condition \(y(0)=y_{0}\) (b) For the data given in Example 1 in the text \([ \leftr=0.71 \text { per year, } K=80.5 \times 10^{6} \mathrm{kg}\), \right. \(\left.y_{0} / K=0.25\right]\), use the Gompertz model to find the predicted value of \(y(2) .\) (c) For the same data as in part (b), use the Gompertz model to find the time \(\tau\) at which \(y(\tau)=0.75 K .\) Hint: You may wish to let \(u=\ln (y / K)\).

Use the technique discussed in Problem 20 to show that the approximation obtained by the Euler method converges to the exact solution at any fixed point as \(h \rightarrow 0 .\) $$ y^{\prime}=\frac{1}{2}-t+2 y, \quad y(0)=1 \quad \text { Hint: } y_{1}=(1+2 h)+t_{1} / 2 $$

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