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Chapter 2: Problem 7

Solve the given differential equation. $$ \frac{d y}{d x}=\frac{x-e^{-x}}{y+e^{y}} $$

Short Answer

Expert verified
Question: Given the differential equation $$ \frac{dy}{dx}=\frac{x-e^{-x}}{y+e^{y}}, $$ find the implicit solution for y(x). Answer: The implicit solution for y(x) is given by $$ \frac{1}{2}y^2 + e^{y} = \frac{1}{2}x^2 + e^{-x} + C. $$

Step by step solution

01

Separate Variables

To separate the variables, we want to get all the y's on one side and all the x's on the other side of the equation. To do this, we can multiply both sides by the denominator of the right side and by dx: $$ (y+e^{y})dy = (x-e^{-x})dx $$ Now we have successfully separated the x's and y's into two separate expressions.
02

Integrate Both Sides

Now, we will integrate both sides of the equation with respect to their respective variables: $$ \int (y+e^{y})dy = \int (x-e^{-x})dx $$ First, integrate the left side with respect to y: $$ \int y dy + \int e^{y} dy = \frac{1}{2}y^2 + e^y + C_1 $$ Next, integrate the right side with respect to x: $$ \int x dx - \int e^{-x} dx = \frac{1}{2}x^2 + e^{-x} + C_2 $$
03

Combine Constants and Solve for y

Now that both sides have been integrated, we will combine the constants and rewrite the equation. We can do this by setting C = C_2 - C_1, and the equation becomes: $$ \frac{1}{2}y^2 + e^y = \frac{1}{2}x^2 + e^{-x} + C $$ To solve for y, we can express y as a function of x. However, due to the complexity, this can't be done explicitly. So the solution to the differential equation can be written implicitly as: $$ \frac{1}{2}y^2 + e^{y} = \frac{1}{2}x^2 + e^{-x} + C $$ The student should note the final solution represents a family of solutions, where C is the constant of integration and can represent any value, which is typical for first-order differential equations.

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Key Concepts

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

Separation of Variables
Solving differential equations can often be a challenging task, but one widely used method to simplify the process is the separation of variables. Separation of variables is a technique where you rearrange a differential equation, which involves a function and its derivative, in such a way that all terms involving one variable are on one side of the equation, and all terms involving the other variable are on the other side.

Imagine you have a dance where participants are initially mixed; separation of variables is like asking all dancers in red to move to one side of the room and all dancers in blue to the other side. In mathematical terms, you manipulate the equation to have one variable, let's say 'y', and its derivatives 'dy' on one side, and the other variable, 'x', and its derivatives 'dx' on the other side. By doing this, you essentially set the stage to solve the equation more straightforwardly through integration.

In the provided exercise, the separation of variables step effectively paves the way to find the solution by isolating y-terms with 'dy' and x-terms with 'dx'. You can view this as sorting like terms into their respective 'camps' to deal with them individually.
Integration of Functions
Once we've successfully separated our variables, the next step in solving the differential equation is to perform integration on both sides. Integration is the process of finding the function that originally produced our derivatives. If differentiation is the process of cutting the birthday cake (our function) into smaller pieces (derivatives), then integration is putting those pieces back together to recreate the whole cake.

In our example, after separation, we have two integrals: one with respect to y and the other with respect to x. By integrating both sides, we're basically summing up the small 'slices' of change outlined by the differential equation to find the original 'whole' functions. This process brings us a step closer to the solution, as it provides us with an equation that reflects the cumulative effect of those changes over the function's entire range.

Calculating the Integrals

Integrating the function on the left side with respect to y and the function on the right side with respect to x yields two separate antiderivatives. Don't forget to add the constant of integration on both sides, since indefinite integrals always include an arbitrary constant, representing the infinite possible positions where our original function could have started its journey.
Implicit Solutions
In the realm of differential equations, not every problem will allow for a neatly packaged explicit solution where the dependent variable is isolated on one side of the equation. This is where implicit solutions come in handy. An implicit solution is a way of defining a function indirectly through an equation in which the function and its derivatives are intermingled.

Think of it as being handed a treasure map with 'X' marking the spot, but the path to the treasure isn't drawn for you. You have to understand the landscape and the clues given to find the treasure. In our exercise, we end up with an equation where 'y' cannot be easily isolated on one side due to the complexity of the relationship between 'y' and 'x'. We solve as much as we can and then represent our answer with both variables interrelated.

The implicit solution is frequently the stopping point for many first-order differential equations. It acknowledges the relationship between x and y without forcibly separating them, allowing for a broader understanding of their connection. In the case of our example, the implicit solution looks like a statement of balance or equivalence between y's and x's behavior, held together by the integration constant 'C', which represents the various potential starting points of our function.

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Most popular questions from this chapter

(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 \(u=\ln (y / K)\).

let \(\phi_{0}(t)=0\) and use the method of successive approximations to solve the given initial value problem. (a) Determine \(\phi_{n}(t)\) for an arbitrary value of \(n .\) (b) Plot \(\phi_{n}(t)\) for \(n=1, \ldots, 4\). Observe whether the iterates appear to be converging. (c) Express \(\lim _{n \rightarrow \infty} \phi_{n}(t)=\phi(t)\) in terms of elementary functions; that is, solve the given initial value problem. (d) Plot \(\left|\phi(t)-\phi_{n}(t)\right|\) for \(n=1, \ldots, 4 .\) For each of \(\phi_{1}(t), \ldots . \phi_{4}(t)\) estimate the interval in which it is a reasonably good approximation to the actual solution. $$ y^{\prime}=-y / 2+t, \quad y(0)=0 $$

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. At a given level of effort, it is reasonable to assume that the rate at which fish are caught depends on the population \(y:\) The more fish there are, the easier it is to catch them. Thus we assume that the rate at which fish are caught is given by \(E y,\) where \(E\) is a positive constant, with units of \(1 /\) time, that measures the total effort made to harvest the given species of fish. To include this effect, the logistic equation is replaced by $$ d y / d t=r(1-y / K) y-E y $$ This equation is known as the Schaefer model after the biologist, M. B. Schaefer, who applied it to fish populations. (a) Show that if \(E0 .\) (b) Show that \(y=y_{1}\) is unstable and \(y=y_{2}\) is asymptotically stable. (c) A sustainable yield \(Y\) of the fishery is a rate at which fish can be caught indefinitely. It is the product of the effort \(E\) and the asymptotically stable population \(y_{2} .\) Find \(Y\) as a function of the effort \(E ;\) the graph of this function is known as the yield-effort curve. (d) Determine \(E\) so as to maximize \(Y\) and thereby find the maximum sustainable yield \(Y_{m}\).

Consider a cylindrical water tank of constant cross section \(A .\) Water is pumped into the tank at a constant rate \(k\) and leaks out through a small hole of area \(a\) in the bottom of the tank. From Torricelli's theorem in hydrodynamics it follows that the rate at which water flows through the hole is \(\alpha a \sqrt{2 g h},\) where \(h\) is the current depth of water in the tank, \(g\) is the acceleration due to gravity, and \(\alpha\) is a contraction coefficient that satisfies \(0.5 \leq \alpha \leq 1.0 .\) (a) Show that the depth of water in the tank at any time satisfies the equation \(-\) $$ d h / d t=(k-\alpha a \sqrt{2 g h}) / A . $$ (b) Determine the equilibrium depth \(h_{e}\) of water and show that it it is asymptotically stable. Observe that \(h_{e}\) does not depend on \(A .\)

Another equation that has been used to model population growth is the Gompertz equation: $$ d y / d t=r y \ln (K / y) $$ where \(r\) and \(K\) are positive constants. (a) Sketch the graph of \(f(y)\) versus \(y,\) find the critical points, and determine whether each is asymptotically stable or unstable. (b) For \(0 \leq y \leq K\) determine where the graph of \(y\) versus \(t\) is concave up and where it is concave down. (c) For each \(y\) in \(0
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