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Chapter 3: Problem 45

The auxiliary equation is \(m(m-1)(m-2)(m-3)+6 m(m-1)(m-2)+3 m(m-1)-3 m+4=0,\) so that \(m_{1}=m_{2}=\sqrt{2}\) and \(m_{3}=m_{4}=-\sqrt{2} .\) The general solution of the differential equation is $$y=c_{1} x^{\sqrt{2}}+c_{2} x^{\sqrt{2}} \ln x+c_{3} x^{-\sqrt{2}}+c_{4} x^{-\sqrt{2}} \ln x.$$

Short Answer

Expert verified
The general solution is: \(y = c_1 x^{\sqrt{2}} + c_2 x^{\sqrt{2}} \ln x + c_3 x^{-\sqrt{2}} + c_4 x^{-\sqrt{2}} \ln x.\)

Step by step solution

01

Understand the Given Equation Root Multiplicity

The auxiliary equation has roots with multiplicity. The roots given are \(m_1 = m_2 = \sqrt{2}\) and \(m_3 = m_4 = -\sqrt{2}\). This means both roots \(\sqrt{2}\) and \(-\sqrt{2}\) appear twice in the equation.
02

Double Roots in Differential Equations

For a double root solution \(m\), the general solution for the differential equation includes terms \(x^m\) and \(x^m \ln x\). In this equation, both \(\sqrt{2}\) and \(-\sqrt{2}\) are double roots.
03

Construct the General Solution

Based on the double root information, the general solution is constructed by adding terms corresponding to each of the roots. This results in four terms: two terms for \(\sqrt{2}\) (\(x^{\sqrt{2}}\) and \(x^{\sqrt{2}} \ln x\)) and two terms for \(-\sqrt{2}\) (\(x^{-\sqrt{2}}\) and \(x^{-\sqrt{2}} \ln x\)).
04

Write Out the General Solution

The general solution has the form: \[ y = c_1 x^{\sqrt{2}} + c_2 x^{\sqrt{2}} \ln x + c_3 x^{-\sqrt{2}} + c_4 x^{-\sqrt{2}} \ln x \]and it takes into account the multiplicity of each root, expressing the solution with constants \(c_1, c_2, c_3,\) and \(c_4\).

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Key Concepts

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

Auxiliary Equation
In the study of differential equations, an auxiliary equation is a key mathematical tool. It is derived from a differential equation using algebraic manipulations. The auxiliary equation helps determine the roots, which are essential for finding the solution of the differential equation.
The auxiliary equation for the given differential equation is \[m(m-1)(m-2)(m-3)+6m(m-1)(m-2)+3m(m-1)-3m+4=0\]This polynomial equation emerges from applying methods like the characteristic equation or other transformations specific to the type of differential equation you are solving.
Once we solve this auxiliary equation, we uncover the crucial roots needed for constructing the general solution of the original differential equation.
Root Multiplicity
Root multiplicity refers to how many times a particular root appears in the auxiliary equation. For this case, the equation is characterized by roots with multiplicity, which means each root appears more than once.
For example, in our equation, the root \(m_1 = m_2 = \sqrt{2}\) appears twice. Similarly, \(m_3 = m_4 = -\sqrt{2}\) also appears twice. Recognizing the multiplicity of roots is vital since it affects the form of the solution.
When roots have higher multiplicities, they contribute additional terms to the general solution, which differ from those of roots with multiplicity of one.
Double Roots
Double roots occur when a root appears exactly twice in the auxiliary equation. For differential equations, this often means modifying the typical form of the solution.
When you encounter double roots, such as \(\sqrt{2}\) and \(-\sqrt{2}\) in our example, the general solution of the differential equation includes two terms for each double root:
  • For \(\sqrt{2}\), the terms are \(x^{\sqrt{2}}\) and \(x^{\sqrt{2}} \ln x\).
  • For \(-\sqrt{2}\), the terms are \(x^{-\sqrt{2}}\) and \(x^{-\sqrt{2}} \ln x\).
This pattern reflects how the presence of double roots enhances the complexity of the solution compared to a simple root.
General Solution
The general solution synthesizes all components derived from solving the auxiliary equation. It involves constructing a combination of terms corresponding to each root and their respective multiplicities.
In our case, the solution integrates the terms from both \(\sqrt{2}\) and \(-\sqrt{2}\) as follows:\[y = c_1 x^{\sqrt{2}} + c_2 x^{\sqrt{2}} \ln x + c_3 x^{-\sqrt{2}} + c_4 x^{-\sqrt{2}} \ln x\]This formulation accounts for each root's multiplicity. The constants \(c_1, c_2, c_3,\) and \(c_4\) represent arbitrary values that can be determined based on initial or boundary conditions given in a specific problem.
The structure of the general solution illustrates the dependence on both the auxiliary equation and the characteristic behavior of its roots, providing a powerful strategy for solving differential equations.

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Most popular questions from this chapter

(a) Setting \(d y / d t=v,\) the differential equation in (13) becomes \(d v / d t=-g R^{2} / y^{2} .\) But, by the chain rule, \(d v / d t=(d v / d y)(d y / d t)=v d v / d t,\) so \(v d v / d y=-g R^{2} / y^{2} .\) Separating variables and integrating we obtain \(v d v=-g R^{2} \frac{d y}{y^{2}} \quad\) and \(\quad \frac{1}{2} v^{2}=\frac{g R^{2}}{y}+c\) Setting \(v=v_{0}\) and \(y=R\) we find \(c=-g R+\frac{1}{2} v_{0}^{2}\) and \\[ v^{2}=2 g \frac{R^{2}}{y}-2 g R+v_{0}^{2} \\] (b) As \(y \rightarrow \infty\) we assume that \(v \rightarrow 0^{+}\). Then \(v_{0}^{2}=2 g R\) and \(v_{0}=\sqrt{2 g R}\) (c) Using \(g=32 \mathrm{ft} / \mathrm{s}\) and \(R=4000(5280) \mathrm{ft}\) we find \(v_{0}=\sqrt{2(32)(4000)(5280)} \approx 36765.2 \mathrm{ft} / \mathrm{s} \approx 25067 \mathrm{mi} / \mathrm{hr}\) (d) \(v_{0}=\sqrt{2(0.165)(32)(1080)} \approx 7760 \mathrm{ft} / \mathrm{s} \approx 5291 \mathrm{mi} / \mathrm{hr}\)

The auxiliary equation is \(m^{2}-9=0,\) so \(y_{c}=c_{1} e^{3 x}+c_{2} e^{-3 x}\) and $$W=\left|\begin{array}{cc}e^{3 x} & e^{-3 x} \\\3 e^{3 x} & -3 e^{-3 x}\end{array}\right|=-6$$ Identifying \(f(x)=9 x / e^{3 x}\) we obtain \(u_{1}^{\prime}=\frac{3}{2} x e^{-6 x}\) and \(u_{2}^{\prime}=-\frac{3}{2} x .\) Then $$\begin{aligned}&u_{1}=-\frac{1}{24} e^{-6 x}-\frac{1}{4} x e^{-6 x}\\\&u_{2}=-\frac{3}{4} x^{2}\end{aligned}$$ and $$\begin{aligned} y &=c_{1} e^{3 x}+c_{2} e^{-3 x}-\frac{1}{24} e^{-3 x}-\frac{1}{4} x e^{-3 x}-\frac{3}{4} x^{2} e^{-3 x} \\\&=c_{1} e^{3 x}+c_{3} e^{-3 x}-\frac{1}{4} x e^{-3 x}(1-3 x).\end{aligned}$$

The thinner curve is obtained using a numerical solver, while the thicker curve is the graph of the Taylor polynomial. We look for a solution of the form \\[ y(x)=y(0)+y^{\prime}(0) x+\frac{1}{2 !} y^{\prime \prime}(0) x^{2}+\frac{1}{3 !} y^{\prime \prime \prime}(0) x^{3}+\frac{1}{4 !} y^{(4)}(0) x^{4}+\frac{1}{5 !} y^{(5)}(0) x^{5} \\] From \(y^{\prime \prime}(x)=x+y^{2}\) we compute \\[ \begin{aligned} y^{\prime \prime \prime}(x) &=1+2 y y^{\prime} \\ y^{(4)}(x) &=2 y y^{\prime \prime}+2\left(y^{\prime}\right)^{2} \\ y^{(5)}(x) &=2 y y^{\prime \prime \prime}+6 y^{\prime} y^{\prime \prime} \end{aligned} \\] Using \(y(0)=1\) and \(y^{\prime}(0)=1\) we find \\[ y^{\prime \prime}(0)=1, \quad y^{\prime \prime \prime}(0)=3, \quad y^{(4)}(0)=4, \quad y^{(5)}(0)=12 \\] An approximate solution is \\[ y(x)=1+x+\frac{1}{2} x^{2}+\frac{1}{2} x^{3}+\frac{1}{6} x^{4}+\frac{1}{10} x^{5} \\]

From \(D x=2 x-y\) and \(D y=x\) we obtain \(y=2 x-D x, D y=2 D x-D^{2} x,\) and \(\left(D^{2}-2 D+1\right) x=0 .\) The solution is $$\begin{aligned} &x=c_{1} e^{t}+c_{2} t e^{t}\\\ &y=\left(c_{1}-c_{2}\right) e^{t}+c_{2} t e^{t} \end{aligned}.$$

The complementary function is \(y_{c}=e^{2 x}\left(c_{1} \cos 2 x+c_{2} \sin 2 x\right) .\) We assume a particular solution of the form \(y_{p}=\left(A x^{3}+B x^{2}+C x\right) e^{2 x} \cos 2 x+\left(D x^{3}+E x^{2}+F\right) e^{2 x} \sin 2 x .\) Substituting into the differential equation and using a CAS to simplify yields $$\left[12 D x^{2}+(6 A+8 E) x+(2 B+4 F)\right] e^{2 x} \cos 2 x$$ $$+\left[-12 A x^{2}+(-8 B+6 D) x+(-4 C+2 E)\right] e^{2 x} \sin 2 x$$ $$=\left(2 x^{2}-3 x\right) e^{2 x} \cos 2 x+\left(10 x^{2}-x-1\right) e^{2 x} \sin 2 x$$ This gives the system of equations $$\begin{array}{ccc} 12 D=2, & 6 A+8 E=-3, & 2 B+4 F=0 \\ -12 A=10, & -8 B+6 D=-1, & -4 C+2 E=-1 \end{array}$$ from which we find \(A=-\frac{5}{6}, B=\frac{1}{4}, C=\frac{3}{8}, D=\frac{1}{6}, E=\frac{1}{4},\) and \(F=-\frac{1}{8} .\) Thus, a particular solution of the differential equation is $$y_{p}=\left(-\frac{5}{6} x^{3}+\frac{1}{4} x^{2}+\frac{3}{8} x\right) e^{2 x} \cos 2 x+\left(\frac{1}{6} x^{3}+\frac{1}{4} x^{2}-\frac{1}{8} x\right) e^{2 x} \sin 2 x$$
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