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Chapter 31: Problem 22

Solve the differential equations below. Find the general solution. (a) \(\frac{d y}{d t}=\sin 3 t\) (b) \(\frac{d y}{d t}=5 \cdot 2^{t}\) (c) \(\frac{d x}{d t}=\frac{t+1}{l}\) (d) \(\frac{d x}{d t}=\frac{t+1}{t^{2}}\)

Short Answer

Expert verified
The general solutions are (a) \(y(t) = -\frac{1}{3}\cos(3t) + C\), (b) \(y(t) = \frac{5 \cdot 2^{t}}{\ln(2)} + C\), (c) \(x(t) = \frac{t^2}{2l} +\frac{t}{l} + C\), and (d) \(x(t) = -\frac{1}{t} + \ln |t| + C\).

Step by step solution

01

Solve equation (a)

The equation \(\frac{d y}{d t}=\sin 3 t\) can be integrated directly. The integral of the right-hand side will give the function \(y(t)\). This yields \(y(t) = -\frac{1}{3}\cos(3t) + C\), where \(C\) is the constant of integration.
02

Solve equation (b)

For the differential equation \(\frac{d y}{d t}=5 \cdot 2^{t}\), you use the rule of exponential integration. The integral of \(b^x\) with respect to \(x\) is \(\frac{b^x}{\ln(b)}\). Thus, \(y(t) = \frac{5\cdot 2^{t}}{\ln(2)} + C\).
03

Solve equation (c)

The differential equation \(\frac{d x}{d t}=\frac{t+1}{l}\) can be integrated directly. By doing so, you get \(x(t) = \frac{t^2}{2l} +\frac{t}{l} + C\).
04

Solve equation (d)

The last differential equation \(\frac{d x}{d t}=\frac{t+1}{t^{2}}\) is a bit more challenging having to use partial fraction decomposition method first. Simplifying, \(\frac{1}{t^2} + \frac{1}{t}\), and then integrating separately gives \(x(t) = -\frac{1}{t} + \ln |t| + C\).

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Key Concepts

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

Integral Calculus
Integral calculus is all about finding the integral of functions, which gives us the area under the curve of a graph of a function. It's the opposite of differentiation. Consider the formula for integration: if you have a function \( f(t) \) and you find its integral, you're essentially reversing the process of finding the derivative.
  • To integrate, constants like \( C \) come into play, known as the constant of integration. They account for any vertical shifts in the function.
  • For example, integrating \( \sin(3t) \) results in \( y(t) = -\frac{1}{3}\cos(3t) + C \).
Understanding integral calculus is crucial for solving differential equations and allows us to explore quantities like areas, volumes, and other quantities that accumulate over time.
Exponential Functions
Exponential functions are characterized by their constant rate of growth; they appear in various fields such as biology, finance, and physics. The general form of an exponential function is \( a \cdot b^x \), where \( a \) is a constant and \( b \) is the base of the exponential.
  • In the context of differential equations, exponential functions arise naturally when modeling growth processes. For example, \( \frac{d y}{d t}=5 \cdot 2^{t} \) is an exponential function.
  • The integral of an exponential function \( b^x \) with respect to \( x \) is \( \frac{b^x}{\ln(b)} + C \).
Recognizing how exponential functions behave and how to integrate them is essential for solving various real-world problems.
Partial Fraction Decomposition
Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions, making them easier to integrate. This method is especially useful when dealing with fractions that have quadratic or higher degree polynomials in the denominator.
  • Take, for instance, the expression \( \frac{t+1}{t^2} \). By breaking it down, it simplifies to \( \frac{1}{t^2} + \frac{1}{t} \).
  • This allows us to integrate each term separately, simplifying the integration process into manageable steps.
Mastering partial fraction decomposition is valuable for tackling complex integrals and obtaining solutions efficiently.
Constant of Integration
When integrating, you often add a constant known as the constant of integration, denoted by \( C \). This constant reflects the fact that an indefinite integral represents a family of functions, not just a single solution.
  • The constant \( C \) accounts for vertical shifts in the graph of the integrated function, ensuring all possible solutions are considered.
  • For example, when integrating \( \sin(3t) \), the output is \( -\frac{1}{3}\cos(3t) + C \).
Understanding the role of the constant of integration is crucial for comprehending the general solution to a differential equation.

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Most popular questions from this chapter

Let \(x=x(t)\) be the number of thousands of animals of species \(A\) at time \(t\). Let \(y=y(t)\) be the number of thousands of animals of species \(B\) at time \(t\). Suppose \(\left\\{\begin{array}{l}\frac{d x}{d t}=x-0.5 x y \\ \frac{d y}{d t}=y-0.5 x y .\end{array}\right.\) (a) Is the interaction between species \(A\) and \(B\) symbiotic, competitive, or a predatorprey relationship? (b) What are the equilibrium populations? (c) Find the nullclines and draw directed horizontal and vertical tangent lines in the phase-plane (as in Figures \(31.28\) and 31.30). (d) The nullclines divide the first quadrant of the phase-plane into four regions. In each region determine the general direction of the trajectories. (e) If \(x=0\), what happens to \(y(t) ?\) How is this indicated in the phase- plane? If \(y=0\), what happens to \(x(t) ?\) How is this indicated in the phase- plane? (f) Use the information gathered in parts (b) through (e) to sketch representative solution trajectories in the phase-plane. Include arrows indicating the direction the trajectories are traveled. (g) For each of the initial conditions given below, describe how the number of species of \(A\) and \(B\) change with time and what the situation will look like in the long run. i. \(x(0)=2 \quad y(0)=1.8\) ii. \(x(0)=2 \quad y(0)=2.3\) iii. \(x(0)=2.2 \quad y(0)=2\) (h) Does this particular model support or challenge Charles Darwin's principle of competitive exclusion?

The population of wildebeest in the Serengeti was decimated by a rinderpest plague in the \(1950 \mathrm{~s}\). In 1961 the Serengeti supported a population of a quarter of a million wildebeest. By 1978 the wildebeest population was \(1.5\) million and by 1991 it had reached 2 million. (Craig Packer Into Africa, Chicago, The University of Chicago Press, 1996 p. \(250 .\) ) Given this data, would you be more inclined to model the growth of the wildebeest population using an exponential growth model or using a logistic growth model? Explain your reasoning.

Sketch a representative family of solutions for each of the following differential equations. (a) \(\frac{d y}{d t}=t^{2}\) (b) \(\frac{d y}{d t}=y^{2}\)

Find the particular solution corresponding to the initial conditions given. \(\frac{d^{2} x}{d t^{2}}+\frac{d x}{d t}=2 x, \quad x(0)=-1, \quad x^{\prime}(0)=0\)

In the beginning of a chemical reaction there are 600 moles of substance \(\mathrm{A}\) and none of substance \(\mathrm{B}\). Over the course of the reaction, the 600 moles of substance \(\mathrm{A}\) are converted to 600 moles of substance B. (Each molecule of A is converted to a molecule of \(\mathrm{B}\) via the reaction.) Suppose the rate at which \(\mathrm{A}\) is turning into \(\mathrm{B}\) is proportional to the product of the number of moles of \(\mathrm{A}\) and the number of moles of \(\mathrm{B}\). (a) Let \(N=N(t)\) be the number of moles of substance \(\mathrm{A}\) at time \(t .\) Translate the statement above into mathematical language. (Note: The number of moles of substance B should be expressed in terms of the number of moles of substance A.) (b) Using your answer to part (a), nd \(\frac{d^{2} N}{d t^{2}}\). Your answer will involve the proportionality constant used in part (a). (c) \(N(t)\) is a decreasing function. The rate at which \(N\) is changing is a function of \(N\), the number of moles of substance A. When the rate at which \(\mathrm{A}\) is being converted to \(\mathrm{B}\) is highest, how many moles are there of substance \(\mathrm{A}\) ?
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