/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 Using a CAS to solve the auxilia... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 59

Using a CAS to solve the auxiliary equation \(3.15 m^{4}-5.34 m^{2}+6.33 m-2.03=0\) we find \(m_{1}=-1.74806\) \(m_{2}=0.501219, m_{3}=0.62342+0.588965 i,\) and \(m_{4}=0.62342-0.588965 i .\) The general solution is $$y=c_{1} e^{-1.74806 x}+c_{2} e^{0.501219 x}+e^{0.62342 x}\left(c_{3} \cos 0.588965 x+c_{4} \sin 0.588965 x\right)$$

Short Answer

Expert verified
The general solution is a combination of exponential and trigonometric functions based on the roots.

Step by step solution

01

Identify the Type of Solution

The given equation is a fourth-degree polynomial in the form \(3.15 m^4 - 5.34 m^2 + 6.33 m - 2.03 = 0\). The complex conjugate roots \(m_3 = 0.62342 + 0.588965i\) and \(m_4 = 0.62342 - 0.588965i\) indicate that the solution will involve exponential functions and trigonometric functions due to the imaginary components.
02

Express Solutions for Real Roots

For the real roots \(m_1 = -1.74806\) and \(m_2 = 0.501219\), the corresponding parts of the general solution are \(c_1 e^{-1.74806 x}\) and \(c_2 e^{0.501219 x}\). These terms are standard exponential solutions for real roots.
03

Formulate Solutions for Complex Roots

For the complex roots \(m_3 = 0.62342 + 0.588965i\) and \(m_4 = 0.62342 - 0.588965i\), we use the form \(e^{ ext{Re}(m)x} (c_3 \, ext{cos}( ext{Im}(m)x) + c_4 \, ext{sin}( ext{Im}(m)x)) \), where \( ext{Re}(m) = 0.62342\) and \( ext{Im}(m) = 0.588965\). Substitute these values to get \(e^{0.62342 x} (c_3 \cos 0.588965 x + c_4 \sin 0.588965 x)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Complex Roots
Complex roots arise when dealing with polynomial equations that cannot be solved solely with real numbers. When a polynomial has roots in the form of \(a + bi\), where \(i\) represents the imaginary unit defined as \(\sqrt{-1}\), those roots are called complex roots. It is essential to note that they often come in conjugate pairs like \(a + bi\) and \(a - bi\).When solving polynomial equations, if you encounter a complex conjugate pair, the solution typically involves combining exponential functions with trigonometric functions. This combination stems from Euler's formula, which relates complex exponentials to sine and cosine functions. Using this approach helps express solutions in terms applicable for practical interpretations in fields like physics and engineering, where real phenomena often involve oscillations and waveforms represented by these functions.
Exponential Functions
Exponential functions are a significant component of the solutions to polynomial equations, particularly when solving for roots. Functions of the form \(e^{ax}\), where \(e\) represents Euler's number (approximately 2.71828), describe how changes occur at a constant rate. This property makes them particularly useful for representing natural processes like radioactive decay or interest compounding in financial models.In the context of solving polynomial equations, exponential functions describe solutions associated with real roots. As seen in our example, the solution consists of exponential terms, \(c_1 e^{-1.74806 x}\) and \(c_2 e^{0.501219 x}\), that link directly to the real parts of the roots found. These functions' roles extend significantly further, supporting the complex parts of solutions involving oscillations when paired with trigonometric functions.
Trigonometric Functions
Trigonometric functions are often used in combination with exponential functions to express solutions to polynomial equations that include complex roots. This use is rooted in their ability to aptly describe periodic phenomena, such as those found in sound waves or alternating current circuits.When dealing with complex roots, as indicated in the exercise, the solution term \(e^{a x}(c_3 \cos(b x) + c_4 \sin(b x))\) emerges, where \(a\) and \(b\) derive from the real and imaginary parts of the complex root. Here, the real part \(a = 0.62342\) lends itself to the exponential, while the imaginary part \(b = 0.588965\) defines the frequency within the trigonometric functions.This blend allows the polynomial's solutions to manifest as visualizable and practically applicable formats, often needed in the realms where frequencies and oscillations are studied, applying well beyond mathematics and into the physical sciences.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For \(\lambda \leq 0\) the only solution of the boundary-value problem is \(y=0 .\) For \(\lambda=\alpha^{2}>0\) we have $$y=c_{1} \cos \alpha x+c_{2} \sin \alpha x$$ Now \(y(-\pi)=y(\pi)=0\) implies \\[ \begin{array}{l} c_{1} \cos \alpha \pi-c_{2} \sin \alpha \pi=0 \\ c_{1} \cos \alpha \pi+c_{2} \sin \alpha \pi=0 \end{array} \\] This homogeneous system will have a nontrivial solution when \\[ \left|\begin{array}{rr} \cos \alpha \pi & -\sin \alpha \pi \\ \cos \alpha \pi & \sin \alpha \pi \end{array}\right|=2 \sin \alpha \pi \cos \alpha \pi=\sin 2 \alpha \pi=0 \\] Then \\[ 2 \alpha \pi=n \pi \quad \text { or } \quad \lambda=\alpha^{2}=\frac{n^{2}}{4} ; \quad n=1,2,3, \ldots \\] When \(n=2 k-1\) is odd, the eigenvalues are \((2 k-1)^{2} / 4 .\) since \(\cos (2 k-1) \pi / 2=0\) and \(\sin (2 k-1) \pi / 2 \neq 0\) we see from either equation in (1) that \(c_{2}=0 .\) Thus, the eigenfunctions corresponding to the eigenvalues \((2 k-1)^{2} / 4\) are \(y=\cos (2 k-1) x / 2\) for \(k=1,2,3, \ldots .\) Similarly, when \(n=2 k\) is even, the eigenvalues are \(k^{2}\) with corresponding eigenfunctions \(y=\sin k x\) for \(k=1,2,3, \ldots\)

Let \((x, y)\) be the coordinates of \(S_{2}\) on the curve \(C .\) The slope at \((x, y)\) is then \\[ d y / d x=\left(v_{1} t-y\right) /(0-x)=\left(y-v_{1} t\right) / x \\] or \(x y^{\prime}-y=-v_{1} t\) Differentiating with respect to \(x\) and using \(r=v_{1} / v_{2}\) gives \\[ \begin{aligned} x y^{\prime \prime}+y^{\prime}-y^{\prime} &=-v_{1} \frac{d t}{d x} \\ x y^{\prime \prime} &=-v_{1} \frac{d t}{d s} \frac{d s}{d x} \\ x y^{\prime \prime} &=-v_{1} \frac{1}{v_{2}}(-\sqrt{1+\left(y^{\prime}\right)^{2}}) \\ x y^{\prime \prime} &=r \sqrt{1+\left(y^{\prime}\right)^{2}} \end{aligned} \\] Letting \(u=y^{\prime}\) and separating variables, we obtain \\[ \begin{aligned} x \frac{d u}{d x} &=r \sqrt{1+u^{2}} \\ \frac{d u}{\sqrt{1+u^{2}}} &=\frac{r}{x} d x \\ \sinh ^{-1} u &=r \ln x+\ln c=\ln \left(c x^{r}\right) \\ u &=\sinh \left(\ln c x^{r}\right) \\ \frac{d y}{d x} &=\frac{1}{2}\left(c x^{r}-\frac{1}{c x^{r}}\right) \end{aligned} \\] At \(t=0, d y / d x=0\) and \(x=a,\) so \(0=c a^{r}-1 / c a^{r} .\) Thus \(c=1 / a^{r}\) and \\[ \frac{d y}{d x}=\frac{1}{2}\left[\left(\frac{x}{a}\right)^{r}-\left(\frac{a}{x}\right)^{r}\right]=\frac{1}{2}\left[\left(\frac{x}{a}\right)^{r}-\left(\frac{x}{a}\right)^{-r}\right] \\] If \(r>1\) or \(r<1,\) integrating gives \\[ y=\frac{a}{2}\left[\frac{1}{1+r}\left(\frac{x}{a}\right)^{1+r}-\frac{1}{1-r}\left(\frac{x}{a}\right)^{1-r}\right]+c_{1} \\] When \(t=0, y=0\) and \(x=a,\) so \(0=(a / 2)[1 /(1+r)-1 /(1-r)]+c_{1} .\) Thus \(c_{1}=a r /\left(1-r^{2}\right)\) and \\[ y=\frac{a}{2}\left[\frac{1}{1+r}\left(\frac{x}{a}\right)^{1+r}-\frac{1}{1-r}\left(\frac{x}{a}\right)^{1-r}\right]+\frac{a r}{1-r^{2}} \\] To see if the paths ever intersect we first note that if \(r>1\), then \(v_{1}>v_{2}\) and \(y \rightarrow \infty\) as \(x \rightarrow 0^{+} .\) In other words, \(S_{2}\) always lags behind \(S_{1}\). Next, if \(r<1\), then \(v_{1}

For \(x^{2} y^{\prime \prime}=0\) the auxiliary equation is \(m(m-1)=0\) and the general solution is \(y=c_{1}+c_{2} x .\) The initial conditions imply \(c_{1}=y_{0}\) and \(c_{2}=y_{1},\) so \(y=y_{0}+y_{1} x .\) The initial conditions are satisfied for all real values of \(y_{0}\) and \(y_{1}\). For \(x^{2} y^{\prime \prime}-2 x y^{\prime}+2 y=0\) the auxiliary equation is \(m^{2}-3 m+2=(m-1)(m-2)=0\) and the general solution is \(y=c_{1} x+c_{2} x^{2}\). The initial condition \(y(0)=y_{0}\) implies \(0=y_{0}\) and the condition \(y^{\prime}(0)=y_{1}\) implies \(c_{1}=y_{1}\). Thus, the initial conditions are satisfied for \(y_{0}=0\) and for all real values of \(y_{1}\). For \(x^{2} y^{\prime \prime}-4 x y^{\prime}+6 y=0\) the auxiliary equation is \(m^{2}-5 m+6=(m-2)(m-3)=0\) and the general solution is \(y=c_{1} x^{2}+c_{2} x^{3}\). The initial conditions imply \(y(0)=0=y_{0}\) and \(y^{\prime}(0)=0 .\) Thus, the initial conditions are satisfied only for \(y_{0}=y_{1}=0\).

From \(k_{1}=40\) and \(k_{2}=120\) we compute the effective spring constant \(k=4(40)(120) / 160=120 .\) Now, \(m=20 / 32\) so \(k / m=120(32) / 20=192\) and \(x^{\prime \prime}+192 x=0 . \quad\) Using \(x(0)=0\) and \(x^{\prime}(0)=2\) we obtain \(x(t)=\frac{\sqrt{3}}{12} \sin 8 \sqrt{3} t\).

Write the equation in the form $$y^{\prime \prime}+\frac{1}{x} y^{\prime}+\left(1-\frac{1}{4 x^{2}}\right) y=x^{-1 / 2}$$ and identify \(f(x)=x^{-1 / 2} .\) From \(y_{1}=x^{-1 / 2} \cos x\) and \(y_{2}=x^{-1 / 2} \sin x\) we compute $$W\left(y_{1}, y_{2}\right)=\left|\begin{array}{cc} x^{-1 / 2} \cos x & x^{-1 / 2} \sin x \\\\-x^{-1 / 2} \sin x-\frac{1}{2} x^{-3 / 2} \cos x & x^{-1 / 2} \cos x-\frac{1}{2} x^{-3 / 2} \sin x\end{array}\right|=\frac{1}{x}$$ Now $$u_{1}^{\prime}=-\sin x \quad \text { so } \quad u_{1}=\cos x,$$ and $$u_{2}^{\prime}=\cos x \quad \text { so } \quad u_{2}=\sin x.$$ Thus a particular solution is $$y_{p}=x^{-1 / 2} \cos ^{2} x+x^{-1 / 2} \sin ^{2} x,$$ and the general solution is $$\begin{aligned}y &=c_{1} x^{-1 / 2} \cos x+c_{2} x^{-1 / 2} \sin x+x^{-1 / 2} \cos ^{2} x+x^{-1 / 2} \sin ^{2} x \\\&=c_{1} x^{-1 / 2} \cos x+c_{2} x^{-1 / 2} \sin x+x^{-1 / 2}.\end{aligned}$$
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Fields in Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Kinematics Physics

Read Explanation

Space Physics

Read Explanation

Famous Physicists

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.