/*! 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 60 Using a CAS to solve the auxilia... [FREE SOLUTION] | 91Ó°ÊÓ

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Using a CAS to solve the auxiliary equation \(m^{4}+2 m^{2}-m+2=0\) we find \(m_{1}=1 / 2+\sqrt{3} i / 2, m_{2}=1 / 2-\sqrt{3} i / 2\) \(m_{3}=-1 / 2+\sqrt{7} i / 2,\) and \(m_{4}=-1 / 2-\sqrt{7} i / 2 .\) The general solution is $$y=e^{x / 2}\left(c_{1} \cos \frac{\sqrt{3}}{2} x+c_{2} \sin \frac{\sqrt{3}}{2} x\right)+e^{-x / 2}\left(c_{3} \cos \frac{\sqrt{7}}{2} x+c_{4} \sin \frac{\sqrt{7}}{2} x\right)$$

Short Answer

Expert verified
The general solution is derived using the roots as coefficients in exponential, cosine, and sine terms.

Step by step solution

01

Understand the Auxiliary Equation

We start with the fourth-degree polynomial equation: \(m^4 + 2m^2 - m + 2 = 0\). We need to find the roots of this polynomial to form the general solution of a differential equation.
02

Use CAS to Find Roots

Using a Computer Algebra System (CAS) to solve the polynomial equation, we obtain four complex roots: \(m_1 = \frac{1}{2} + \frac{\sqrt{3}}{2} i\), \(m_2 = \frac{1}{2} - \frac{\sqrt{3}}{2} i\), \(m_3 = -\frac{1}{2} + \frac{\sqrt{7}}{2} i\), and \(m_4 = -\frac{1}{2} - \frac{\sqrt{7}}{2} i\). These roots are in the form of \(a \pm bi\).
03

Formulate the General Solution

For each pair of complex conjugate roots of the form \(a \pm bi\), the corresponding part of the general solution is \(e^{ax}(C_1 \cos(bx) + C_2 \sin(bx))\). Apply this to both pairs: \(m_1, m_2\) and \(m_3, m_4\).
04

Construct the Full Solution Using Both Pairs of Roots

1. For roots \(m_1\) and \(m_2\):- The solution component is \(e^{\frac{x}{2}} (c_1 \cos \frac{\sqrt{3}}{2}x + c_2 \sin \frac{\sqrt{3}}{2}x)\).2. For roots \(m_3\) and \(m_4\):- The solution component is \(e^{-\frac{x}{2}} (c_3 \cos \frac{\sqrt{7}}{2}x + c_4 \sin \frac{\sqrt{7}}{2}x)\).
05

Combine Components into the General Solution

Putting both components together, we derive the general solution:\[y = e^{\frac{x}{2}}(c_1 \cos \frac{\sqrt{3}}{2}x + c_2 \sin \frac{\sqrt{3}}{2}x) + e^{-\frac{x}{2}}(c_3 \cos \frac{\sqrt{7}}{2}x + c_4 \sin \frac{\sqrt{7}}{2}x)\]

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Key Concepts

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

Auxiliary Equation
An auxiliary equation is a key concept in understanding differential equations. It's particularly important for solving linear differential equations with constant coefficients. The purpose of finding the auxiliary equation is to determine the characteristic roots that will help us form the general solution. In our case, we started with the polynomial equation \(m^4 + 2m^2 - m + 2 = 0\). This polynomial serves as the auxiliary equation. By finding its roots, we pave the way to construct the general solution for the given differential equation. Remember, solving this polynomial is crucial because the solutions (roots) will directly influence how we express the general solution.
General Solution
The general solution of a differential equation represents a family of solutions that include arbitrary constants. These constants allow the solution to be customized based on initial or boundary conditions. For differential equations with constant coefficients, once we have the characteristic roots from the auxiliary equation, constructing the general solution becomes straightforward. If the roots are real, the solution involves exponential terms. However, if they are complex, as in our case, the solution elegantly incorporates both exponential and trigonometric functions. In our example, the general solution is formed by combining terms for each pair of complex conjugate roots. This includes
  • \(e^{x/2}(c_1 \cos \frac{\sqrt{3}}{2} x + c_2 \sin \frac{\sqrt{3}}{2} x)\)
  • \(e^{-x/2}(c_3 \cos \frac{\sqrt{7}}{2} x + c_4 \sin \frac{\sqrt{7}}{2} x)\)
These terms collectively form the complete general solution.
Complex Roots
Complex roots are a common occurrence when solving higher-degree polynomials, especially in the context of differential equations. When dealing with complex roots, they usually appear in conjugate pairs of the form \(a \pm bi\). Here, \(a\) and \(b\) are real numbers, while \(i\) represents the imaginary unit. In our scenario, the auxiliary equation yields four complex roots:
  • \(m_1 = \frac{1}{2} + \frac{\sqrt{3}}{2} i\)
  • \(m_2 = \frac{1}{2} - \frac{\sqrt{3}}{2} i\)
  • \(m_3 = -\frac{1}{2} + \frac{\sqrt{7}}{2} i\)
  • \(m_4 = -\frac{1}{2} - \frac{\sqrt{7}}{2} i\)
To form the general solution with complex roots, we make use of Euler's formula, which helps convert complex exponential functions into combinations of sine and cosine functions. This transformation results in the beautiful and functional trigonometric expressions seen in the general solution.
Computer Algebra System (CAS)
A Computer Algebra System, or CAS, is a tool that greatly aids in solving complicated equations that are often cumbersome to handle manually. CAS can efficiently perform symbolic mathematics, such as factorizing large polynomials or computing complex roots. In this exercise, a CAS was instrumental in solving the polynomial \(m^4 + 2m^2 - m + 2 = 0\). Given the complexity of the equation, finding the roots accurately might be challenging without such a tool. CAS quickly identifies the roots of the polynomial, providing both real and complex results with ease. Using CAS, therefore, not only saves time and reduces potential manual errors but also allows for a deeper focus on understanding and constructing solutions, rather than being bogged down by lengthy algebraic manipulations.

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Most popular questions from this chapter

Solving \(\frac{1}{2} q^{\prime \prime}+20 q^{\prime}+1000 q=0\) we obtain \(q_{c}(t)=e^{-20 t}\left(c_{1} \cos 40 t+c_{2} \sin 40 t\right) .\) The steady- state charge has the form \(q_{p}(t)=A \sin 60 t+B \cos 60 t+C \sin 40 t+D \cos 40 t .\) Substituting into the differential equation we find $$\begin{aligned}(-1600 A-2400 B) & \sin 60 t+(2400 A-1600 B) \cos 60 t \\ +&(400 C-1600 D) \sin 40 t+(1600 C+400 D) \cos 40 t\end{aligned}$$ $$=200 \sin 60 t+400 \cos 40 t$$ Equating coefficients we obtain \(A=-1 / 26, B=-3 / 52, C=4 / 17,\) and \(D=1 / 17 .\) The steady-state charge is \\[q_{p}(t)=-\frac{1}{26} \sin 60 t-\frac{3}{52} \cos 60 t+\frac{4}{17} \sin 40 t+\frac{1}{17} \cos 40 t\\] and the steady-state current is \\[i_{p}(t)=-\frac{30}{13} \cos 60 t+\frac{45}{13} \sin 60 t+\frac{160}{17} \cos 40 t-\frac{40}{17} \sin 40 t\\].

Using the general solution \(y=c_{1} \cos \sqrt{3} x+c_{2} \sin \sqrt{3} x+2 x,\) the boundary conditions \(y(0)+y^{\prime}(0)=0, y(1)=0\) yield the system $$\begin{array}{r} c_{1}+\sqrt{3} c_{2}+2=0 \\ c_{1} \cos \sqrt{3}+c_{2} \sin \sqrt{3}+2=0 \end{array}$$ Solving gives $$c_{1}=\frac{2(-\sqrt{3}+\sin \sqrt{3})}{\sqrt{3} \cos \sqrt{3}-\sin \sqrt{3}} \quad \text { and } \quad c_{2}=\frac{2(1-\cos \sqrt{3})}{\sqrt{3} \cos \sqrt{3}-\sin \sqrt{3}}$$ Thus, $$y=\frac{2(-\sqrt{3}+\sin \sqrt{3}) \cos \sqrt{3} x}{\sqrt{3} \cos \sqrt{3}-\sin \sqrt{3}}+\frac{2(1-\cos \sqrt{3}) \sin \sqrt{3} x}{\sqrt{3} \cos \sqrt{3}-\sin \sqrt{3}}+2 x$$

If \(\lambda=\alpha^{2}=P / E I,\) then the solution of the differential equation is \\[ y=c_{1} \cos \alpha x+c_{2} \sin \alpha x+c_{3} x+c_{4} \\]. The conditions \(y(0)=0, y^{\prime \prime}(0)=0\) yield, in turn, \(c_{1}+c_{4}=0\) and \(c_{1}=0 .\) With \(c_{1}=0\) and \(c_{4}=0\) the solution is \(y=c_{2} \sin \alpha x+c_{3} x .\) The conditions \(y(L)=0, y^{\prime \prime}(L)=0,\) then yield $$c_{2} \sin \alpha L+c_{3} L=0 \quad \text { and } \quad c_{2} \sin \alpha L=0.$$Hence, nontrivial solutions of the problem exist only if \(\sin \alpha L=0 .\) From this point on, the analysis is the same as in Example 3 in the text.

We have \(y_{c}=c_{1} \cosh x+c_{2} \sinh x\) and we assume \(y_{p}=A x \cosh x+B x \sinh x .\) Substituting into the differential equation we find \(A=0\) and \(B=\frac{1}{2} .\) Thus $$y=c_{1} \cosh x+c_{2} \sinh x+\frac{1}{2} x \sinh x$$ From the initial conditions we obtain \(c_{1}=2\) and \(c_{2}=12,\) so $$y=2 \cosh x+12 \sinh x+\frac{1}{2} x \sinh x$$

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 $$\begin{aligned}y &=c_{1} x^{-1 / 2} \cos x+c_{2} x^{-1 / 2} \sin x+x^{-1 / 2} \cos ^{2} x+x^{-1 / 2} \sin ^{2} x \\\&=c_{1} x^{-1 / 2} \cos x+c_{2} x^{-1 / 2} \sin x+x^{-1 / 2}.\end{aligned}$$

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