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Chapter 5: Problem 44

Solve the initial-value problems. (a) \(\frac{d y}{d x}=\frac{1}{(2 x)^{3}}, y(1)=0\) (b) \(\frac{d y}{d t}=\sec ^{2} t-\sin t, y\left(\frac{\pi}{4}\right)=1\) (c) \(\frac{d y}{d x}=x^{2} \sqrt{x^{3}}, y(0)=0\)

Short Answer

Expert verified
(a) \( y = -\frac{1}{16x^2} + \frac{1}{16} \); (b) \( y = \tan t + \cos t - \frac{\sqrt{2}}{2} \); (c) \( y = \frac{2}{9} x^{9/2} \)."

Step by step solution

01

Solve the differential equation for part (a)

We start with the differential equation \( \frac{d y}{d x} = \frac{1}{(2x)^3} \). To solve for \( y \), we integrate both sides with respect to \( x \). First, simplify \( \frac{1}{(2x)^3} = \frac{1}{8x^3} \). Now integrate: \[ y = \int \frac{1}{8x^3} \, dx = \frac{1}{8} \int x^{-3} \, dx. \] The integral of \( x^{-3} \) is \( -\frac{1}{2}x^{-2} \), so: \[ y = \frac{1}{8} \left(-\frac{1}{2}x^{-2}\right) = -\frac{1}{16x^2} + C. \]
02

Apply the initial condition for part (a)

The initial condition is \( y(1) = 0 \). Using the solution from Step 1, substitute \( x = 1 \) and \( y = 0 \): \[ 0 = -\frac{1}{16(1)^2} + C \] This simplifies to \( C = \frac{1}{16} \). Thus, the specific solution is: \[ y = -\frac{1}{16x^2} + \frac{1}{16}. \]
03

Solve the differential equation for part (b)

For the equation \( \frac{d y}{d t} = \sec^2 t - \sin t \), integrate to find \( y(t) \). Calculate the integrals: \( \int \sec^2 t \, dt = \tan t \) and \( \int \sin t \, dt = -\cos t \). So, \[ y(t) = \tan t + \cos t + C. \]
04

Apply the initial condition for part (b)

The initial condition is \( y\left(\frac{\pi}{4}\right) = 1 \). Substitute \( t = \frac{\pi}{4} \) and \( y = 1 \): \[ 1 = \tan\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) + C \] Since \( \tan\left(\frac{\pi}{4}\right) = 1 \) and \( \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \), \[ 1 = 1 + \frac{\sqrt{2}}{2} + C. \] Therefore, \( C = 1 - 1 - \frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{2} \). Thus, the specific solution is: \[ y(t) = \tan t + \cos t - \frac{\sqrt{2}}{2}. \]
05

Solve the differential equation for part (c)

The differential equation is \( \frac{d y}{d x} = x^2 \sqrt{x^3} \). The term \( \sqrt{x^3} = x^{3/2} \), so rewrite the equation as \( \frac{d y}{d x} = x^{2 + 3/2} = x^{7/2} \). Integrate: \[ y = \int x^{7/2} \, dx = \frac{2}{9} x^{9/2} + C. \]
06

Apply the initial condition for part (c)

The initial condition is \( y(0) = 0 \). Substitute \( x = 0 \) and \( y = 0 \) into the general solution: \[ 0 = \frac{2}{9} (0)^{9/2} + C \] Since \( \frac{2}{9} (0)^{9/2} = 0 \), we have \( C = 0 \). Thus, the specific solution is: \[ y(x) = \frac{2}{9} x^{9/2}. \]

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Key Concepts

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

Differential Equations
A differential equation involves an unknown function and its derivatives. In the context of the given problems, we deal with first-order differential equations, which include expressions like \( \frac{d y}{d x} \) or \( \frac{d y}{d t} \). These represent the rate of change of the dependent variable, \( y \), with respect to an independent variable, say \( x \) or \( t \). Solving these equations means finding an expression for \( y \) that satisfies the given condition. Sometimes, we may need to simplify an expression before integrating, as shown in the solution steps. This process often involves rewriting terms to make integration manageable.
Integration
Integration is a core technique in solving differential equations. It involves finding the antiderivative or the original function from its derivative. In this exercise, we see integration being used to solve equations like \( \frac{d y}{d x} = \frac{1}{8x^3} \), where we find \( y \) by integrating both sides with respect to \( x \).
  • The integration of \( x^{-3} \) results in \( -\frac{1}{2}x^{-2} \), illustrating how we find such antiderivatives.
  • In the second example, integration with respect to \( t \) involves trigonometric identities: \( \int \sec^2 t \, dt = \tan t \) and \( \int \sin t \, dt = -\cos t \).
Through integration, we derive a general solution, which then gets refined based on initial conditions.
Initial Conditions
Initial conditions are specific values provided for an equation that allow us to find the particular solution of a differential equation. They are crucial because they help determine the constant of integration, \( C \), which arises after performing an indefinite integral.
  • For part (a), the condition \( y(1) = 0 \) helps find \( C \) by plugging in the values \( x = 1 \) and \( y = 0 \) into the equation \( y = -\frac{1}{16x^2} + C \).
  • In part (b), \( y\left(\frac{\pi}{4}\right) = 1 \) allows us to solve for \( C \) in \( y(t) = \tan t + \cos t + C \).
This particularization makes the solution unique to the problem's initial conditions.
Calculus Concepts
Some fundamental calculus concepts appear throughout these solutions, linking derivatives, integrals, and other mathematical techniques. Understanding these allows students to grasp the intricacies of solving differential equations with initial conditions.
  • Antiderivatives: Finding the original function from its derivative, exemplified in solving \( \frac{d y}{d x} \).
  • Trigonometric Integration: Applying knowledge of trigonometric identities and integration, such as \( \tan t \) and \( \cos t \) integrals.
  • Substitution in Initial Conditions: Utilizing given values of \( y \) and \( x \) or \( t \) to determine constants.
These concepts help form the building blocks for tackling more complex calculus problems.

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Most popular questions from this chapter

(a) Over what open interval does the formula $$ F(x)=\int_{1}^{x} \frac{1}{t^{2}-9} d t $$ represent an antiderivative of $$ f(x)=\frac{1}{x^{2}-9} ? $$ (b) Find a point where the graph of \(F\) crosses the \(x\) -axis.

Use a calculating utility to find the midpoint approximation of the integral using \(n=20\) sub-intervals, and then find the exact value of the integral using Part 1 of the Fundamental Theorem of Calculus. $$\int_{0}^{\pi / 2} \sin x d x$$

Evaluate the integrals using appropriate substitutions. $$\int \frac{d x}{\sqrt{x} e^{(2 \sqrt{x})}}$$

Electricity is supplied to homes in the form of alternating current, which means that the voltage has a sinusoidal waveform described by an equation of the form $$V=V_{p} \sin (2 \pi f t)$$ (see the accompanying figure). In this equation, \(V_{p}\) is called the peak voltage or amplitude of the current, \(f\) is called its frequency, and \(1 / f\) is called its period. The voltages \(V\) and \(V_{p}\) are measured in volts (V), the time \(t\) is measured in seconds (s), and the frequency is measured in hertz (Hz). (1 \(\mathrm{Hz}=1\) cycle per second; a cycle is the electrical term for one period of the waveform.) Most alternating-current voltmeters read what is called the rms or root -mean-square value of \(V .\) By definition, this is the square root of the average value of \(V^{2}\) over one period. (a) Show that $$V_{\mathrm{rms}}=\frac{V_{p}}{\sqrt{2}}$$ [Hint: Compute the average over the cycle from \(t=0\) to \(t=1 / f,\) and use the identity \(\sin ^{2} \theta=\frac{1}{2}(1-\cos 2 \theta)\) to help evaluate the integral. \(]\) (b) In the United States, electrical outlets supply alternating current with an rms voltage of \(120 \mathrm{V}\) at a frequency of \(60 \mathrm{Hz}\). What is the peak voltage at such an outlet? (Check your book to see figure)

Use a calculating utility to find the midpoint approximation of the integral using \(n=20\) sub-intervals, and then find the exact value of the integral using Part 1 of the Fundamental Theorem of Calculus. $$\int_{-1}^{1} \sec ^{2} x d x$$
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