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Solve the differential equation. $$ \left(e^{y}-1\right) y^{\prime}=2+\cos x $$

Short Answer

Expert verified
The solution is given by \(e^y - y = 2x + \sin x + C\).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \((e^y - 1) y' = 2 + \cos x\). Recognize it as a first-order differential equation involving both \(x\) and \(y\), suitable for the separation of variables method.
02

Re-arrange to Separate Variables

We rewrite the equation as \( (e^y - 1) dy = (2 + \cos x) dx\). This re-arranges the equation to separate the variables \(y\) and \(x\) on each side.
03

Integrate Both Sides

Integrate both sides individually: \(\int (e^y - 1) \, dy = \int (2 + \cos x) \, dx\). The left side becomes \(e^y - y + C_1\) and the right side becomes \(2x + \sin x + C_2\). Simplify by combining constants to get \(e^y - y = 2x + \sin x + C\).
04

Solve for the Constant of Integration

Recognize that \(C\) represents an arbitrary constant that can be determined if an initial condition is given. Without an initial condition, \(C\) remains unsolved.

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Key Concepts

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

Separation of Variables
Separation of variables is a powerful method used to solve first-order differential equations like the one in our problem. This technique involves rearranging the given equation so that all terms involving one variable, say \(y\), are on one side, and all terms involving the other variable, \(x\), are on the other side.

This allows us to deal with each variable independently and integrate them separately, which simplifies solving the equation. In our exercise, this method is used to separate the equation \((e^y - 1) y' = 2 + \cos x\) into separate parts for \(y\) and \(x\) which results in:
  • \((e^y - 1) dy = (2 + \cos x) dx\)
Once separated, we can easily proceed to integrate both sides - a crucial step toward finding the solution. Separating the variables paves the way for the next steps of integration and solving for the unknowns.
First-order Differential Equation
First-order differential equations are the simplest type of differential equations. They involve only the first derivative of the unknown function. In our exercise, the equation given is of the form \( (e^y - 1) y' = 2 + \cos x \), which clearly involves the first derivative \( y' \).

First-order differential equations are often solved by methods such as separation of variables or integrating factors, each useful in different scenarios. Recognizing the order and type of differential equation helps determine which solution method to apply. Identifying our differential equation as first-order allowed us to choose an effective method - separation of variables - to tackle the problem.
  • Includes only the first derivative
  • Can be linear or non-linear
  • Offers various solution techniques
Understanding this classification shapes the entire approach in solving the problem efficiently.
Integration Techniques
Integration techniques are essential in solving differential equations, especially once the variables are separated. In our step-by-step solution, integration plays a critical role after variables \(y\) and \(x\) are segregated.

For the equation \( \int (e^y - 1) \, dy = \int (2 + \cos x) \, dx \), we tackle integration of each side:
  • The left side \( \int (e^y - 1) \, dy \) simplifies to \( e^y - y \)
  • The right side \( \int (2 + \cos x) \, dx \) results in \( 2x + \sin x \)
Mastering basic integration techniques, such as integrating exponential functions or trigonometric functions like \( \cos x \), is crucial. Also, remembering to include the constant of integration \( C \) is important for capturing the general solution. Integration transforms our initial differential equation into a more solvable form, getting us close to the final answer.

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Most popular questions from this chapter

Solve the differential equation. $$ t^{2} \frac{d y}{d t}+3 t y=\sqrt{1+t^{2}}, \quad t>0 $$

Another model for a growth function for a limited population is given by the Gompertz function, which is a solution of the differential equation $$ \frac{d P}{d t}=c \ln \left(\frac{M}{P}\right) P $$ where \(c\) is a constant and \(M\) is the carrying capacity. (a) Solve this differential equation. (b) Compute lim \(_{t \rightarrow \infty} P(t)\). (c) Graph the Gompertz growth function for \(M=1000\), \(P_{0}=100,\) and \(c=0.05,\) and compare it with the logistic function in Example \(2 .\) What are the similarities? What are the differences? (d) We know from Exercise 13 that the logistic function grows fastest when \(P=M / 2\). Use the Gompertz differential equation to show that the Gompertz function grows fastest when \(P=M / e\).

In a seasonal-growth model, a periodic function of time is introduced to account for seasonal variations in the rate of growth. Such variations could, for example, be caused by seasonal changes in the availability of food. (a) Find the solution of the seasonal-growth model $$ \frac{d P}{d t}=k P \cos (r t-\phi) \quad P(0)=P_{0} $$ where \(k, r,\) and \(\phi\) are positive constants. (b) By graphing the solution for several values of \(k, r,\) and \(\phi,\) explain how the values of \(k, r,\) and \(\phi\) affect the solution. What can you say about \(\lim _{t \rightarrow \infty} P(t) ?\)

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Lynx eat snowshoe hares and snowshoe hares eat woody plants like willows. Suppose that, in the absence of hares, the willow population will grow exponentially and the lynx population will decay exponentially. In the absence of lynx and willow the hare population will decay exponentially. If \(L(t), H(t),\) and \(W(t)\) represent the populations of these three species at time \(t\) write a system of differential equations as a model for their dynamics. If the constants in your equation are all positive, explain why you have used plus or minus signs.

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