/*! 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 6 From \(m^{2}-8 m+20=0\) we find ... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 6

From \(m^{2}-8 m+20=0\) we find \(m_{1}=4+2 i\) and \(m_{2}=4-2 i .\) Then \(y_{c}=e^{4 x}\left(c_{1} \cos 2 x+c_{2} \sin 2 x\right)\) and we assume \(y_{p}=A x^{2}+B x+C+(D x+E) e^{x} .\) Substituting into the differential equation we obtain $$\begin{aligned} 2 A-8 B+20 C &=0 \\ -6 D+13 E &=0 \\ -16 A+20 B &=0 \\ 13 D &=-26 \\ 20 A &=100 \end{aligned}$$ Then \(A=5, B=4, C=\frac{11}{10}, D=-2, E=-\frac{12}{13}, y_{p}=5 x^{2}+4 x+\frac{11}{10}+\left(-2 x-\frac{12}{13}\right) e^{x}\) and \(y=e^{4 x}\left(c_{1} \cos 2 x+c_{2} \sin 2 x\right)+5 x^{2}+4 x+\frac{11}{10}+\left(-2 x-\frac{12}{13}\right) e^{x}\)

Short Answer

Expert verified
The solution is \( y = e^{4x}(c_1 \cos 2x + c_2 \sin 2x) + 5x^2 + 4x + \frac{11}{10} + (-2x - \frac{12}{13})e^x \).

Step by step solution

01

Identify Known Values

We know the roots of the characteristic equation are \( m_1 = 4 + 2i \) and \( m_2 = 4 - 2i \). The corresponding complementary solution is \( y_c = e^{4x} (c_1 \cos 2x + c_2 \sin 2x) \).
02

Set Up the Particular Solution

Assume a particular solution of the form \( y_p = Ax^2 + Bx + C + (Dx + E)e^x \) to account for polynomial and exponential terms.
03

Substitute into Differential Equation

We substitute terms and equate coefficients for terms \( Ax^2, Bx, C, Dx, E \) due to the differential equation system. This yields the given system of equations after matching coefficients.
04

Solve Equation System

Solve the simultaneous equations obtained from substitution:\[\begin{aligned}20A &= 100, & \Rightarrow A = 5 \-16A + 20B &= 0, & \Rightarrow B = 4 \2A - 8B + 20C &= 0, & \Rightarrow C = \frac{11}{10} \13D &= -26, & \Rightarrow D = -2 \-6D + 13E &= 0, & \Rightarrow E = -\frac{12}{13}\end{aligned}\]
05

Formulate the Final Solution

Combine the complementary solution and the particular solution to get the overall solution:\[ y = e^{4x}(c_1 \cos 2x + c_2 \sin 2x) + 5x^2 + 4x + \frac{11}{10} + (-2x - \frac{12}{13})e^x \]

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Key Concepts

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

Particular Solution
A particular solution is a specific solution to a non-homogeneous differential equation. It satisfies the equation by including terms that compensate for the non-homogeneous part of the equation.

The goal in finding a particular solution is to determine the coefficients of the assumed form that accurately fit the equation. For our example, we presumed a particular solution of the form \( y_p = Ax^2 + Bx + C + (Dx + E)e^x \). This assumption aimed to match the equation's polynomial and exponential terms.

By substituting this assumed form into the differential equation, we established a system of equations to find the correct values of \( A, B, C, D, \) and \( E \). This step might seem complex, but it involves simple algebra鈥攅quating coefficients of like terms, leading us to our particular solution.
Complementary Solution
The complementary solution is an essential part of solving linear differential equations, especially the homogeneous ones. It specifically addresses the homogeneous equation related to the main differential equation.

In our particular example, the complementary solution was derived from solving the characteristic equation. It resulted in the solution: \( y_c = e^{4x}(c_1 \cos 2x + c_2 \sin 2x) \).

Characteristic roots help determine the complementary solution, which often involves exponential and trigonometric terms. These terms contribute to each of the modes of the system being analyzed and must fit within the homogeneous part of the differential equation. Complementary solutions are crucial as they describe the natural behavior of the system without external forces.
Characteristic Equation
The characteristic equation is a crucial concept in finding solutions to linear differential equations, especially second-order ones. It's derived from setting the associated homogeneous equation to zero.

In our discussed example, the characteristic equation was found to be \( m^2 - 8m + 20 = 0 \). Solving this quadratic equation gives the characteristic roots, \( m_1 = 4 + 2i \) and \( m_2 = 4 - 2i \).

These roots are complex, indicating the possibility of oscillatory behavior, which is conveyed in the complementary solution using sine and cosine functions. Understanding the characteristic equation and its roots is vital as they directly influence the form of the complementary solution. It's a foundational step to comprehending how the overall solution behaves.

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Most popular questions from this chapter

From the solution \(y_{1}=e^{-4 x} \cos x\) we conclude that \(m_{1}=-4+i\) and \(m_{2}=-4-i\) are roots of the auxiliary equation. Hence another solution must be \(y_{2}=e^{-4 x} \sin x .\) Now dividing the polynomial \(m^{3}+6 m^{2}+m-34\) by \([m-(-4+i)][m-(-4-i)]=m^{2}+8 m+17\) gives \(m-2 .\) Therefore \(m_{3}=2\) is the third root of the auxiliary equation, and the general solution of the differential equation is $$y=c_{1} e^{-4 x} \cos x+c_{2} e^{-4 x} \sin x+c_{3} e^{2 x}$$

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 \(y^{\prime \prime}(x)=x^{2}+y^{2}-2 y^{\prime}\) we compute \\[ \begin{aligned} y^{\prime \prime \prime}(x) &=2 x+2 y y^{\prime}-2 y^{\prime \prime} \\ y^{(4)}(x) &=2+2\left(y^{\prime}\right)^{2}+2 y y^{\prime \prime}-2 y^{\prime \prime \prime} \\ y^{(5)}(x) &=6 y^{\prime} y^{\prime \prime}+2 y y^{\prime \prime \prime}-2 y^{(4)} \end{aligned} \\] Using \(y(0)=1\) and \(y^{\prime}(0)=1\) we find \\[ y^{\prime \prime}(0)=-1, \quad y^{\prime \prime \prime}(0)=4, \quad y^{(4)}(0)=-6, \quad y^{(5)}(0)=14 \\] An approximate solution is \\[ y(x)=1+x-\frac{1}{2} x^{2}+\frac{2}{3} x^{3}-\frac{1}{4} x^{4}+\frac{7}{60} x^{5} \\]

The auxiliary equation is \(m(m-1)(m-2)+4 m(m-1)+5 m-9=0,\) so that \(m_{1}=1.40819\) and the two complex roots are \(-1.20409 \pm 2.22291 i .\) The general solution of the differential equation is $$y=c_{1} x^{1.40819}+x^{-1.20409}\left[c_{2} \cos (2.22291 \ln x)+c_{3} \sin (2.22291 \ln x)\right].$$

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 $$\begin{aligned} y &=c_{1} e^{3 x}+c_{2} e^{-3 x}-\frac{1}{24} e^{-3 x}-\frac{1}{4} x e^{-3 x}-\frac{3}{4} x^{2} e^{-3 x} \\\&=c_{1} e^{3 x}+c_{3} e^{-3 x}-\frac{1}{4} x e^{-3 x}(1-3 x).\end{aligned}$$

(a) The weight of \(x\) feet of the chain is \(2 x,\) so the corresponding mass is \(m=2 x / 32=x / 16 .\) The only force acting on the chain is the weight of the portion of the chain hanging over the edge of the platform. Thus, by Newton's second law, \(\frac{d}{d t}(m v)=\frac{d}{d t}\left(\frac{x}{16} v\right)=\frac{1}{16}\left(x \frac{d v}{d t}+v \frac{d x}{d t}\right)=\frac{1}{16}\left(x \frac{d v}{d t}+v^{2}\right)=2 x\) and \(x d v / d t+v^{2}=32 x .\) Now, by the Chain Rule, \(d v / d t=(d v / d x)(d x / d t)=v d v / d x,\) so \(x v d v / d x+v^{2}=32 x\) (b) We separate variables and write the differential equation as \(\left(v^{2}-32 x\right) d x+x v d v=0 .\) This is not an exact form, but \(\mu(x)=x\) is an integrating factor. Multiplying by \(x\) we get \(\left(x v^{2}-32 x^{2}\right) d x+x^{2} v d v=0 .\) This form is the total differential of \(u=\frac{1}{2} x^{2} v^{2}-\frac{32}{3} x^{3},\) so an implicit solution is \(\frac{1}{2} x^{2} v^{2}-\frac{32}{3} x^{3}=c .\) Letting \(x=3\) and \(v=0\) we find \(c=-288 .\) Solving for \(v\) we get \(\frac{d x}{d t}=v=\frac{8 \sqrt{x^{3}-27}}{\sqrt{3} x}, \quad 3 \leq x \leq 8\) (c) Separating variables and integrating we obtain \(\frac{x}{\sqrt{x^{3}-27}} d x=\frac{8}{\sqrt{3}} d t \quad\) and \(\quad \int_{3}^{x} \frac{s}{\sqrt{s^{3}-27}} d s=\frac{8}{\sqrt{3}} t+c\) since \(x=3\) when \(t=0,\) we see that \(c=0\) and \\[ t=\frac{\sqrt{3}}{8} \int_{3}^{x} \frac{s}{\sqrt{s^{3}-27}} d s \\] We want to find \(t\) when \(x=7 .\) Using a CAS we find \(t(7)=0.576\) seconds.
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