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Chapter 8: Problem 7

Solve each pure-time differential equation. \(\frac{d s}{d t}=\sqrt{3 t+1}\), where \(s(0)=1\)

Short Answer

Expert verified
The particular solution to the differential equation is \( s = \frac{2}{9} (3t + 1)^{3/2} + \frac{7}{9} \).

Step by step solution

01

Separate Variables

First, we separate the variables to express the equation in terms of differentials. This gives us the equation \[ ds = \sqrt{3t + 1} \, dt. \]
02

Integrate Both Sides

Integrate both sides of the equation with respect to their respective variables. We have \[ \int ds = \int \sqrt{3t + 1} \, dt. \] The left-hand side integrates to \( s \). The right-hand side requires a substitution: let \( u = 3t + 1 \), so \( du = 3 \, dt \) or \( dt = \frac{1}{3} du \). Substitute to get \[ \int \sqrt{u} \times \frac{1}{3} \, du = \frac{1}{3} \int u^{1/2} \, du. \]
03

Solve the Integral of the Right-Hand Side

Solve the integral \( \int u^{1/2} \, du \) which is a power rule integral giving us \[ \frac{1}{3} \cdot \frac{2}{3} u^{3/2} = \frac{2}{9} u^{3/2}. \] Substitute back \( u = 3t + 1 \) to get \[ \frac{2}{9} (3t + 1)^{3/2}. \]
04

Write the General Solution

The general solution to the differential equation is \[ s = \frac{2}{9} (3t + 1)^{3/2} + C, \] where \( C \) is the constant of integration.
05

Apply Initial Condition

Use the initial condition \( s(0) = 1 \) to find \( C \). Substituting \( t = 0 \) and \( s = 1 \) into the equation gives \[ 1 = \frac{2}{9} (3 \times 0 + 1)^{3/2} + C. \]Simplify to find \[ 1 = \frac{2}{9} \cdot 1^{3/2} + C \rightarrow 1 = \frac{2}{9} + C. \] Solve for \( C \): \[ C = 1 - \frac{2}{9} = \frac{7}{9}. \]
06

Write the Particular Solution

Substitute \( C = \frac{7}{9} \) into the general solution: \[ s = \frac{2}{9} (3t + 1)^{3/2} + \frac{7}{9}. \] This is the particular solution that satisfies the initial condition.

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Key Concepts

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

Integration
Integration is a fundamental concept in calculus concerned with determining the area under a curve. In the context of differential equations, integration allows us to find the function given its derivative. In simpler terms, it is the reverse process of differentiation.

When solving a differential equation, integrating both sides of the equation is a common step. Take for instance the equation from the original exercise:
  • We had the differential equation in the form \[ \int ds = \int \sqrt{3t + 1} \, dt. \]
  • The left side, \(ds\), integrated to \(s\), which represents the original function we were seeking. This integration essentially tells us what function has a derivative of \(\frac{d s}{d t}\).
  • The right side involved integrating \(\sqrt{3t + 1}\). However, because it's not straightforward, we made use of the substitution method to simplify our task. The substitution \(u = 3t + 1\) made the integration process clearer, allowing us to apply the power rule wonderfully.

The result from this entire process is a general solution to the differential equation, represented mathematically with a constant \(C\), indicating that there are potentially infinite functions that can solve this equation, differing by a constant factor.
Initial Conditions
Initial conditions give us additional information needed to determine a specific solution from a general solution of a differential equation. They represent the values of your dependent variable at particular values of the independent variable.

These conditions are critical because when you solve a differential equation, the integration step introduces an arbitrary constant, \(C\), in the solution. For example:
  • We found the general solution as \[ s = \frac{2}{9} (3t + 1)^{3/2} + C. \]
  • Without initial conditions, \(C\) can be any real number, leading to infinitely many solutions.
    So, the initial condition \(s(0) = 1\) was used. We plugged \(t = 0\) into the general solution \[ 1 = \frac{2}{9} \cdot 1^{3/2} + C \]
  • This equation allowed us to solve for \(C\) and determine a specific solution rather than a family of solutions.

Initial conditions are not only pivotal in narrowing down to a single solution but also in applications such as physics and engineering, where they represent real-world constraints at the starting point of an observation or process.
Variables Separation
Separation of variables is a method used to solve differential equations where you rearrange the equation such that each variable is on a separate side of the equation. This technique is particularly useful for first-order differential equations.

In our specific problem:
  • The differential equation given was \[ \frac{d s}{d t} = \sqrt{3t + 1}. \]
  • To separate variables, we rearranged this into differentials so that terms with \(s\) were on one side and terms with \(t\) were on the other: \[ ds = \sqrt{3t + 1} \, dt. \]
  • This allowed each side to be integrated independently.

By doing this, we transform the complexity of solving a differential equation into two simpler integration problems. This approach works effectively when the multiplicative form of the variables can be separated, thus simplifying the process significantly.

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Most popular questions from this chapter

This problem addresses Malthus's concerns. Assume that a population size grows exponentially according to $$N(t)=1000 e^{t}.$$ and the food supply grows linearly according to $$F(t)=3 t$$ (a) Write a differential equation for each of \(N(t)\) and \(F(t)\). (b) What assumptions do you need to make to be able to compare whether and, if so, when food supply will be insufficient? Does exponential growth eventually overtake linear growth? Explain. (c) Do a Web search to determine whether food supply has grown linearly, as claimed by Malthus. 57\. At the beginning of this section, we modified the exponentialgrowth equation to include oscillations in the per capita growth rate. Solve the differential equation we obtained, namely, $$\frac{d N}{d t}=2(1+\sin (2 \pi t)) N(t)$$ with \(N(0)=5\).

Solve each pure-time differential equation. \(\frac{d y}{d x}=\frac{1}{1+x^{2}}\), where \(y_{0}=1\) when \(x_{0}=0\)

Assume that the size of a population evolves according to the logistic equation with intrinsic rate of growth \(r=1.5\). Assume that the carrying capacity \(K=100\). (a) Find the differential equation that describes the rate of growth of this population. (b) Find all equilibria, and, using the graphical approach, discuss the stability of the equilibria. (c) Find the eigenvalues associated with the equilibria, and use the eigenvalues to determine the stability of the equilibria. Compare your answers with your results in (b).

Denote by \(p=p(t)\) the fraction of occupied patches in a metapopulation model, and assume that $$\frac{d p}{d t}=c p(1-p)-p^{2} \quad \text { for } t \geq 0$$ where \(c>0 .\) The term \(p^{2}\) describes the density-dependent extinction of patches; that is, the per-patch extinction rate is \(p\), and a fraction \(p\) of patches are occupied, resulting in an extinction rate of \(p^{2}\). The colonization of vacant patches is the same as in the Levins model. (a) Set \(g(p)=c p(1-p)-p^{2}\) and sketch the graph of \(g(p)\). (b) Find all equilibria of \((8.70)\) in \([0,1]\), and determine their stability. (c) Is there a nontrivial equilibrium when \(c>0 ?\) Contrast your findings with the corresponding results in the Levins model.

(Adapted from Crawley, 1997) Denote plant biomass by \(V\), and herbivore number by \(N .\) The plant-herbivore interaction is modeled as $$ \begin{array}{l} \frac{d V}{d t}=a V\left(1-\frac{V}{K}\right)-b V N \\ \frac{d N}{d t}=c V N-d N \end{array} $$ (a) Suppose the herbivore number is equal to \(0 .\) What differential equation describes the dynamics of the plant biomass? Can you explain the resulting equation? Determine the plant biomass equilibrium in the absence of herbivores. (b) Now assume that herbivores are present. Describe the effect of herbivores on plant biomass; that is, explain the term \(-b V N\) in the first equation. Describe the dynamics of the herbivoresthat is, how their population size increases and what contributes to decreases in their population size. (c) Determine the equilibria (1) by solving $$ \frac{d V}{d t}=0 \quad \text { and } \quad \frac{d N}{d t}=0 $$ and (2) graphically. Explain why this model implies that "plant abundance is determined solely by attributes of the herbivore," as stated in Crawley (1997).
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