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Chapter 3: Problem 3

In Problems \(1-16\), the indicated function \(y_{1}(x)\) is a solution of the given equation. Use reduction of order or formula (5), as instructed, to find a second solution \(y_{2}(x)\). $$ y^{\prime \prime}+16 y=0 ; \quad y_{1}=\cos 4 x $$

Short Answer

Expert verified
The second solution is \( y_2(x) = \frac{1}{4} \sin(4x) \).

Step by step solution

01

Identify the Given Differential Equation and Known Solution

The given differential equation is \( y'' + 16y = 0 \) and one known solution is \( y_1(x) = \cos(4x) \). Our task is to find a second solution \( y_2(x) \) that is linearly independent of the first solution \( y_1(x) \).
02

Recognize the Homogeneous Linear Differential Equation

This is a second-order linear homogeneous differential equation with constant coefficients. We see that the characteristic equation is used to determine the solutions to this equation.
03

Use the Reduction of Order Formula

According to reduction of order, if one solution \( y_1(x) \) is known, a second linearly independent solution \( y_2(x) \), can be found using the formula: \[ y_2(x) = y_1(x) \int \frac{e^{-\int P(x) \, dx}}{(y_1(x))^2} \, dx \]Here, the equation is of the form \( y'' + P(x)y' + Q(x)y = 0 \), simplified to our case, \( P(x) = 0 \) since the equation is \( y'' + 16y = 0 \).
04

Substitute and Simplify

Substitute into the formula:\[ y_2(x) = \cos(4x) \int \frac{1}{\cos^2(4x)}\, dx \]This simplifies to:\[ y_2(x) = \cos(4x) \int \sec^2(4x)\, dx \]
05

Integrate the Expression

The integral of \( \sec^2(4x) \) is \( \frac{1}{4}\tan(4x) \). Therefore:\[ y_2(x) = \cos(4x) \cdot \frac{1}{4} \tan(4x) \]Upon simplification, this becomes:\[ y_2(x) = \frac{1}{4} \sin(4x) \]
06

Confirm the Linear Independence

The Wronskian of \( y_1(x) = \cos(4x) \) and \( y_2(x) = \frac{1}{4} \sin(4x) \) should be nonzero:\[ W(y_1, y_2) = \begin{vmatrix} \cos(4x) & \frac{1}{4} \sin(4x) \ -4\sin(4x) & \frac{1}{4} \cdot 4\cos(4x) \end{vmatrix} \]This determinant evaluates to \( 1 \), confirming that \( y_2(x) \) is a valid second solution.

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Key Concepts

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

Differential Equation
A differential equation is an equation that involves the derivatives of a function. Simply put, it relates a function to its rate of change. In this exercise, we deal with a type of differential equation known as a second-order linear homogeneous differential equation:

  • **Second-order:** This refers to the highest derivative in the equation, which in this case is the second derivative, denoted as \( y'' \).
  • **Linear:** The equation can be written in the form \( a_2(x)y'' + a_1(x)y' + a_0(x)y = 0 \), where \( a_2(x) \), \( a_1(x) \), and \( a_0(x) \) are coefficients that can be functions of \( x \). Here, our coefficients are constants, making it easier to solve.
  • **Homogeneous:** This tells us that every term in the equation involves the function \( y \) or its derivatives. There is no additional function on the right-hand side. This generally implies that the solution will pass through the origin if \( y = 0 \) is a solution.
These characteristics help determine the methods used for finding solutions, like the characteristic equation and reduction of order.
Linearly Independent Solution
In differential equations, finding linearly independent solutions is crucial. Linearly independent solutions cannot be expressed as a simple multiple of each other. They allow the complete general solution of a differential equation to be constructed.

Here's why linear independence matters:
  • For second-order differential equations like ours, we always look for two linearly independent solutions, \( y_1(x) \) and \( y_2(x) \).
  • These solutions form the basis for the solution space, meaning any other solution to the differential equation can be expressed as a linear combination of these two.
The process we use here, called reduction of order, helps find the second linearly independent solution when one is already known.
Knowing \( y_1(x) = \cos(4x) \), we aim to find \( y_2(x) \) that does not overlap it in the solution space, allowing us to span the entire set of solutions.
Characteristic Equation
The characteristic equation is a tool that simplifies finding solutions to differential equations with constant coefficients. It originates from assuming a solution of the form \( y = e^{rx} \), where \( r \) is a constant we need to solve for.

Here's how it works in our exercise:
  • Given the differential equation \( y'' + 16y = 0 \), we substitute \( y = e^{rx} \), leading to \( r^2e^{rx} + 16e^{rx} = 0 \).
  • This simplifies to the characteristic equation \( r^2 + 16 = 0 \). Solving for \( r \), we find \( r = \pm 4i \), indicating a complex pair.
With complex roots, the general solution is expressed using sine and cosine functions. The initial real solution \( y_1(x) = \cos(4x) \) comes from this. Another complementary solution, typically involving the sine function, can be derived, supporting the problem's use of reduction of order.
Wronskian
The Wronskian is an important determinant used to confirm the linear independence of solutions in differential equations. Here's how it applies:

  • The Wronskian of two functions, \( y_1(x) \) and \( y_2(x) \), is calculated as a determinant: \[ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} \]
  • If the Wronskian is non-zero at any point in the interval of interest, the solutions \( y_1 \) and \( y_2 \) are linearly independent, meaning they form a valid basis for the solution space.
  • For our solutions, \( y_1(x) = \cos(4x) \) and \( y_2(x) = \frac{1}{4} \sin(4x) \), the Wronskian evaluates to 1. This confirms that these two solutions are indeed linearly independent.
Thus, employing the Wronskian ensures that we've obtained a valid set of solutions for our differential equation, helping conclude the solution process reliably.

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Most popular questions from this chapter

$$ 6.11 y^{\prime \prime \prime}+8.59 y^{\prime \prime}+7.93 y^{\prime}+0.778 y=0 $$

Consider a pendulum that is released from rest from an initial displacement of \(\theta_{0}\) radians. Solving the linear model (7) subject to the initial conditions \(\theta(0)=\theta_{0}, \theta^{\prime}(0)=0\) gives \(\theta(t)=\theta_{0} \cos \sqrt{g} \| t .\) Theperiod of oscillations predicted by this modelisgivenbythefamiliarformula \(T=2 \pi / \sqrt{g l l}=2 \pi \sqrt{U g}\). The interesting thing about this formula for \(T\) is that it does not depend on the magnitude of the initial displacement \(\theta_{0}\). In other words, the linear model predicts that the time that it would take the pendulum to swing from an initial displacement of, say, \(\theta_{0}=\pi / 2\left(=90^{\circ}\right)\) to \(-\pi / 2\) and back again would be exactly the same time to cycle from, say, \(\theta_{0}=\pi / 360\left(=0.5^{\circ}\right)\) to \(-\pi / 360\). This is intuitively unreasonable; the actual period must depend on \(\theta_{0}\). If we assume that \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) and \(l=32 \mathrm{ft}\), then the period of oscillation of the linear model is \(T=2 \pi \mathrm{s}\). Let us compare this last number with the period predicted by the nonlinear model when \(\theta_{0}=\pi / 4\). Using a numerical solver that is capable of generating hard data, approximate the solution of $$\frac{d^{2} \theta}{d t^{2}}+\sin \theta=0, \quad \theta(0)=\frac{\pi}{4}, \quad \theta^{\prime}(0)=0$$ for \(0 \leq t \leq 2\). As in Problem 24 , if \(t_{1}\) denotes the first time the pendulum reaches the position \(O P\) in Figure 3.11.3, then the period of the nonlinear pendulum is \(4 t_{1} .\) Here is another way of solving the equation \(\theta(t)=0\). Expeniment with small step sizes and advance the time staning at \(t=0\) and ending at \(t=2\). From your hard data, observe the time \(t_{1}\) when \(\theta(t)\) changes, for the first time, from positive to negative. Use the value \(t_{1}\) to determine the true value of the period of the nonlinear pendulum. Compute the percentage relative error in the period estimated by \(T=2 \pi\).

When the magnitude of tension \(T\) is not constant, then a model for the deflection curve or shape \(y(x)\) assumed by a rotating string is given by $$ \frac{d}{d x}\left[T(x) \frac{d y}{d x}\right]+\rho \omega^{2} y=0 $$ Suppose that \(10.25\), show that the critical speeds of angular rotation are $$ \omega_{n}=\frac{1}{2} \sqrt{\left(4 n^{2} \pi^{2}+1\right) / \rho} $$ and the corresponding deflections are $$ y_{n}(x)=c_{2} x^{-1 / 2} \sin (n \pi \ln x), n=1,2,3, \ldots $$ (b) Use a graphing utility to graph the deflection curves on the interval \([1, e]\) for \(n=1,2,3 .\) Choose \(c_{2}=1\)

In Problems, find the eigenvalues and eigenfunctions for the given boundary- value problem. $$ y^{\prime \prime}+(\lambda+1) y=0, y^{\prime}(0)=0, y^{\prime}(1)=0 $$

$$ \text { Solve the given initial-value problem. } $$ $$ y^{\prime \prime \prime}+2 y^{\prime \prime}-5 y^{\prime}-6 y \quad 0, y(0) \quad y^{\prime}(0) \quad 0, y^{\prime \prime}(0) \quad 1 $$
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