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Chapter 4: Problem 30

Solve the given initial-value problem. $$\frac{d^{2} y}{d \theta^{2}}+y=0, \quad y(\pi / 3)=0, y^{\prime}(\pi / 3)=2$$

Short Answer

Expert verified
The solution is \( y(\theta) = -\sqrt{3} \cos \theta + \sin \theta \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \( \frac{d^{2} y}{d \theta^{2}} + y = 0 \). This is a linear homogeneous second-order differential equation with constant coefficients.
02

Write the Characteristic Equation

For the homogeneous differential equation \( \frac{d^{2} y}{d \theta^{2}} + y = 0 \), we use the characteristic equation \( r^2 + 1 = 0 \).
03

Solve the Characteristic Equation

Solve \( r^2 + 1 = 0 \). This gives \( r = \pm i \) since \( r = \sqrt{-1} \).
04

Write the General Solution

Using the solutions \( r = \pm i \), the general solution is \( y(\theta) = c_1 \cos \theta + c_2 \sin \theta \).
05

Apply Initial Condition \( y(\pi/3) = 0 \)

Substitute \( \theta = \pi/3 \) into the general solution: \( 0 = c_1 \cos(\pi/3) + c_2 \sin(\pi/3) \). This simplifies to \( 0 = \frac{c_1}{2} + \frac{c_2\sqrt{3}}{2} \).
06

Apply Initial Condition \( y'(\pi/3) = 2 \)

Differentiating the general solution gives \( y'(\theta) = -c_1 \sin \theta + c_2 \cos \theta \). Substitute \( \theta = \pi/3 \), which gives \( 2 = -c_1 \sin(\pi/3) + c_2 \cos(\pi/3) \). This simplifies to \( 2 = -\frac{c_1\sqrt{3}}{2} + \frac{c_2}{2} \).
07

Solve the System of Equations

Solve the system of equations from steps 5 and 6: 1. \( \frac{c_1}{2} + \frac{c_2\sqrt{3}}{2} = 0 \)2. \( -\frac{c_1\sqrt{3}}{2} + \frac{c_2}{2} = 2 \).Multiplying equation 1 by 2 gives \( c_1 + c_2\sqrt{3} = 0 \) so \( c_1 = -c_2\sqrt{3} \). Substitute into equation 2 and solve for \( c_2 \) and \( c_1 \).
08

Calculate Constants \( c_1 \) and \( c_2 \)

Substitute \( c_1 = -c_2\sqrt{3} \) into equation 2: \( -\frac{(-c_2\sqrt{3})\sqrt{3}}{2} + \frac{c_2}{2} = 2 \).This becomes \( \frac{3c_2}{2} + \frac{c_2}{2} = 2 \) or \( 2c_2 = 2 \), yielding \( c_2 = 1 \).Substitute \( c_2 = 1 \) back to find \( c_1 = -\sqrt{3} \).
09

Write the Solution with the Constants

Substitute \( c_1 = -\sqrt{3} \) and \( c_2 = 1 \) into the general solution: \( y(\theta) = -\sqrt{3} \cos \theta + \sin \theta \).

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Key Concepts

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

Initial-Value Problem
An initial-value problem in the realm of differential equations is where you need to find a specific solution to a differential equation that satisfies given conditions called initial conditions. These initial conditions are typically specified values of the unknown function and its derivatives at a certain point.
This exercise presents an initial-value problem where the second-order differential equation is given alongside initial conditions: \( y(\pi/3) = 0 \) and \( y'(\pi/3) = 2 \).
  • The first part of solving an initial-value problem involves finding the general solution to the differential equation without accounting for any conditions.
  • Subsequently, you apply the initial conditions to this general solution to find specific constants, thus deriving a particular solution.
By satisfying these conditions, the solution is tailored to the problem's specific scenario.
Characteristic Equation
To solve a differential equation like the one in our problem, we often convert the differential equation into a simpler algebraic form known as the characteristic equation. This step is crucial when solving linear differential equations with constant coefficients.
  • The given equation \( \frac{d^{2} y}{d \theta^{2}} + y = 0 \) translates into the characteristic equation by substituting derivatives with powers of a variable, usually \( r \).
  • In this example, the characteristic equation becomes \( r^2 + 1 = 0 \).
Characteristics equations help identify the nature of the solutions. Here, solving \( r^2 + 1 = 0 \) gives complex roots \( r = \pm i \), indicating solutions that involve trigonometric functions.
Homogeneous Differential Equation
A homogeneous differential equation is one where every term is dependent on the unknown function or its derivatives. There are no standalone terms (constant or explicitly function-free).
In our exercise, \( \frac{d^{2} y}{d \theta^{2}} + y = 0 \) is defined as homogeneous because each term either involves \( y \) or its derivatives. This kind of equation typically has solutions in the form of trigonometric or exponential functions based on the characteristic equation's roots:
  • Real roots lead to exponential functions in the solution.
  • Complex roots, as we encounter here, give solutions involving sine and cosine functions.
For our problem, the trigonometric nature help us derive a general solution in terms of sine and cosine: \( y(\theta) = c_1 \cos\theta + c_2 \sin\theta \).
Second-Order Differential Equation
Second-order differential equations are equations involving the second derivative of a function. They are central in modeling systems with acceleration, such as physical and mechanical systems.
Our problem features the second-order differential equation \( \frac{d^{2} y}{d \theta^{2}} + y = 0 \). Solving second-order equations usually follows a structured method:
  • Identify if the equation is homogeneous or non-homogeneous.
  • Derive the characteristic equation and solve it to find its roots.
  • Formulate the general solution based on those roots.
  • Apply specific initial or boundary conditions to refine the solution to a particular one.
These steps ensure that the obtained solution is robust and correctly aligned with any real-world phenomena the equation represents.

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Most popular questions from this chapter

Solve the given differential equation by undetermined coefficients. $$y^{\prime \prime}+2 y^{\prime}+2 y=5 e^{6 x}$$

Solve the given initial-value problem. $$\begin{aligned} &y^{\prime \prime \prime}-2 y^{\prime \prime}+y^{\prime}=x e^{x}+5, \quad y(0)=2, y^{\prime}(0)=2\\\ &y^{\prime \prime}(0)=-1 \end{aligned}$$

Find a homogeneous linear differential equation with constant coefficients whose general solution is given. $$y=c_{1} e^{-4 x}+c_{2} e^{-3 x}$$

The indicated functions are known linearly independent solutions of the associated homogeneous differential equation on \((0, \infty)\). Find the general solution of the given non-homogeneous equation. $$\begin{aligned} &x^{2} y^{\prime \prime}+x y^{\prime}+y=\sec (\ln x)\\\ &y_{1}=\cos (\ln x), y_{2}=\sin (\ln x) \end{aligned}$$

Consider the differential equation \(a y^{\prime \prime}+b y^{\prime}+c y=e^{k x}\) where \(a, b, c,\) and \(k\) are constants. The auxiliary equation of the associated homogeneous equation is \(a m^{2}+b m+c=0\). (a) If \(k\) is not a root of the auxiliary equation, show that we can find a particular solution of the form \(y_{p}=A e^{k x},\) where \(A=1 /\left(a k^{2}+b k+c\right)\) (b) If \(k\) is a root of the auxiliary equation of multiplicity one, show that we can find a particular solution of the form \(y_{p}=A x e^{k x},\) where \(A=1 /(2 a k+b) .\) Explain how we know that \(k \neq-b /(2 a)\) (c) If \(k\) is a root of the auxiliary equation of multiplicity two, show that we can find a particular solution of the form \(y=A x^{2} e^{k x},\) where \(A=1 /(2 a)\)
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