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Chapter 5: Problem 70

Solve the initial-value problems. $$ \frac{d y}{d x}=2+\sin 3 x, y(\pi / 3)=0 $$

Short Answer

Expert verified
The particular solution is \( y = 2x - \frac{1}{3} \cos 3x - \frac{2\pi}{3} - \frac{1}{3} \).

Step by step solution

01

Understanding the Problem

We need to solve the first-order differential equation \( \frac{d y}{d x} = 2 + \sin 3x \) with the given initial condition \( y(\pi/3) = 0 \).
02

Integrating the Differential Equation

To find the general solution, we integrate both sides of the equation: \[ \int \frac{d y}{d x} \, dx = \int (2 + \sin 3x) \, dx \] Integrating, we get:\[ y = \int 2 \, dx + \int \sin 3x \, dx \] Evaluating these integrals, we obtain:\[ y = 2x - \frac{1}{3} \cos 3x + C \] where \( C \) is the constant of integration.
03

Applying the Initial Condition

Using the initial condition \( y(\pi/3) = 0 \), we substitute into the general solution:\[ 0 = 2 \left(\frac{\pi}{3}\right) - \frac{1}{3} \cos \left(3 \times \frac{\pi}{3}\right) + C \]This simplifies to:\[ 0 = \frac{2\pi}{3} - \frac{1}{3} \cos \pi + C \]We know \( \cos \pi = -1 \), so the equation becomes:\[ 0 = \frac{2\pi}{3} + \frac{1}{3} + C \]Simplifying:\[ C = -\frac{2\pi}{3} - \frac{1}{3} \]
04

Writing the Particular Solution

Substitute \( C \) back into the general solution to obtain the particular solution:\[ y = 2x - \frac{1}{3} \cos 3x - \frac{2\pi}{3} - \frac{1}{3} \]

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Key Concepts

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

First-order Differential Equation
When we talk about a first-order differential equation, we're referring to an equation that involves the derivatives of a function and the function itself.
These types of equations are called "first-order" because they involve the first derivative, or the rate of change, of a function. Here's a simple breakdown of what a first-order differential equation entails:
  • The equation usually looks like \( \frac{dy}{dx} = f(x, y) \), where \( \frac{dy}{dx} \) is the first derivative of \( y \) with respect to \( x \).
  • It shows how the rate of change of \( y \) is related to \( y \) as well as \( x \).
  • To solve such an equation, we often need to find a function \( y \) that satisfies this relationship.
In our exercise, \( \frac{d y}{d x} = 2 + \sin 3x \) is the first-order differential equation we're solving.
We're looking for a function \( y \) that changes over \( x \) in the specific way that this equation describes.
Integrating Differential Equations
Integrating a differential equation is a crucial step in finding a solution. Essentially, integration is the reverse of differentiation, and it helps us "undo" the derivative to find the original function.
In our exercise, we need to integrate the right-hand side of the equation, \( 2 + \sin 3x \), with respect to \( x \) to find \( y \).Let's break it down:
  • Integrating \( 2 \) with respect to \( x \) gives \( 2x \).
  • Integrating \( \sin 3x \) involves a little more work, using the integral formula for sine, \( \int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) \).
  • This results in \( -\frac{1}{3} \cos 3x \) when \( k = 3 \).
  • Don't forget the constant of integration, \( C \), which represents the unknown value that could be added to our solution.
So the integrated function becomes \( y = 2x - \frac{1}{3} \cos 3x + C \), where \( C \) is determined by further conditions.
Initial Condition
An initial condition helps us find a unique solution to a differential equation from among a family of possible solutions.
It's like having a starting point that gives us the specific path the solution needs to follow.Here's why initial conditions are important:
  • Without an initial condition, our solution \( y = 2x - \frac{1}{3} \cos 3x + C \) would have an infinite number of possible solutions because of the constant \( C \).
  • By providing a specific value, like \( y(\pi/3) = 0 \), we can solve for \( C \) and find one particular solution.
  • In our problem, substituting the initial condition into the equation allowed us to find \( C = -\frac{2\pi}{3} - \frac{1}{3} \), ensuring the solution meets the given criteria.
The initial condition makes our solution specific to the conditions of the problem, transforming it from general to particular.

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Most popular questions from this chapter

(a) Let $$I=\int_{0}^{a} \frac{f(x)}{f(x)+f(a-x)} d x$$ Show that \(I=a / 2\). [Hint: Let \(u=a-x,\) and then note the difference between the resulting integrand and \(1 .]\) (b) Use the result of part (a) to find $$\int_{0}^{3} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{3-x}} d x$$ (c) Use the result of part (a) to find $$\int_{0}^{\pi / 2} \frac{\sin x}{\sin x+\cos x} d x$$

Find the area under the curve \(y=9 /(x+2)^{2}\) over the interval \([-1,1] .\)

Use Part 2 of the Fundamental Theorem of Calculus to find the derivatives. $$ \frac{d}{d u} \int_{0}^{u}|x| d x $$

Evaluate the integrals by any method. $$ \int_{1}^{2} \sqrt{5 x-1} d x $$

Evaluate the integrals by any method. $$ \int_{\pi^{2}}^{4 \pi^{2}} \frac{1}{\sqrt{x}} \sin \sqrt{x} d x $$
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