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Chapter 3: Problem 1

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 $$\begin{aligned} &x=c_{1} e^{t}+c_{2} t e^{t}\\\ &y=\left(c_{1}-c_{2}\right) e^{t}+c_{2} t e^{t} \end{aligned}.$$

Short Answer

Expert verified
The solution is \(x = c_1 e^t + c_2 t e^t\) and \(y = (c_1 - c_2)e^t + c_2 t e^t\).

Step by step solution

01

Analyze the Given Equations

We have two differential equations: \(D x = 2x - y\) and \(D y = x\). The goal is to solve these equations to find \(x\) and \(y\).
02

Express y in Terms of x and Derivatives

From the first equation \(y = 2x - Dx\), substitute \(y\) in terms of \(x\) and its derivative in the second equation which gives \(D y = x\).
03

Substitute into Derivative Equations

Substitute \(y = 2x - Dx\) into \(D y = x\), leading to \(D(2x - Dx) = x\), and simplify to form an equation in terms of \(x\).
04

Simplify Derivative Formulation

Differentiate \(y = 2x - Dx\) to get a new equation: \(D y = 2Dx - D^2x\). Substitute \(D y = x\) from earlier steps to find a second equation: \(x = 2Dx - D^2x\).
05

Combine and Rearrange Equations

Combine the equations \(D x = 2x - y\) and \(D y = 2Dx - D^2x\) to form a single second-order differential equation: \((D^2 - 2D + 1)x = 0\).
06

Solve the Differential Equation for x

The characteristic equation of \((D^2 - 2D + 1)x = 0\) is \(r^2 - 2r + 1 = 0\). This factors into \((r - 1)^2 = 0\), giving a repeated root \(r = 1\). The general solution is \(x = c_1 e^t + c_2 t e^t\).
07

Solve for y using the Expression for x

Use the relationship \(y = 2x - Dx\) to express \(y\) in terms of \(t\). Substitute \(x = c_1 e^t + c_2 t e^t\) and differentiate to find \(Dx\), then solve for \(y\). This yields \(y = (c_1 - c_2)e^t + c_2 t e^t\).

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Key Concepts

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

Second-order differential equation
A second-order differential equation is an equation that involves the second derivative of a function. In the context of this problem, we derive a second-order equation from two first-order systems. The road to achieving this involves several steps:
  • First, express one variable in terms of the other and its derivative. This helps in reducing the number of equations.
  • Then, differentiate and rearrange to express the derivative relationships as a single equation involving a second derivative.
Ultimately, the second-order differential equation we come to solve here is \[ (D^2 - 2D + 1)x = 0 \].This is a standard form where the highest derivative is squared (second order). Understanding how to read and manipulate such equations is key as they appear often in physics and engineering to describe dynamic systems.
Repeated roots
Repeated roots occur in solving differential equations when the characteristic equation gives us the same root more than once. Here, the characteristic equation\[ r^2 - 2r + 1 = 0 \]factors into\[ (r - 1)^2 = 0 \],indicating that the root, \( r = 1 \), repeats. This impacts how we formulate the solution:
  • Instead of simple exponential solutions, repeated roots necessitate the use of an additional term \( c_2 t e^{rt} \), to account for the repetition in a unique solution.
  • Thus, the general solution has to include both exponential parts and an extra term for completeness.
Hence, for our problem, the x-solution takes the form:\[ x = c_1 e^t + c_2 t e^t \].This ensures the solution covers all necessary parts of the function's behavior as determined by the repetition found in characteristic roots.
Characteristic equation
The characteristic equation is derived from the differential equation by substituting a trial solution of the form \( e^{rt} \). For a second-order differential equation \[ (D^2 - 2D + 1)x = 0 \],substituting \( x = e^{rt} \) provides:
  • \( D x = r e^{rt} \)
  • \( D^2 x = r^2 e^{rt} \)
Substituting these into the differential equation profers\[ r^2 e^{rt} - 2r e^{rt} + e^{rt} = 0 \].This simplifies to\[ (r^2 - 2r + 1)e^{rt} = 0 \],aligning with the characteristic equation \[ r^2 - 2r + 1 = 0 \].Solving this helps identify the types of roots we have, leading to understanding how the overall solution set looks.
Solution of differential equations
Solving a differential equation involves finding a function or set of functions that satisfy the equation. For second-order linear equations with constant coefficients, such as\[ (D^2 - 2D + 1)x = 0 \],we solve using exponential trial solutions based on the characteristic roots. Once the characteristic roots are determined, the structure of the solution follows:
  • If roots are real and distinct, the solution includes terms \( c_1 e^{r_1 t} + c_2 e^{r_2 t} \).
  • For repeated roots, include an additional term \( c_2 t e^{r t} \) to ensure completeness, as seen in our solution: \( x = c_1 e^t + c_2 t e^t \).
From here, determining associated system variables, like \( y \) in our problem, involves substituting back known solutions and using given relations (e.g., \( y = 2x - Dx \)).This transforms abstract mathematical insights into concrete steps for rebuilding the behaviors and interactions described by the system of differential equations.

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Most popular questions from this chapter

Solving \(\frac{1}{20} q^{\prime \prime}+2 q^{\prime}+100 q=0\) we obtain \(q(t)=e^{-20 t}\left(c_{1} \cos 40 t+c_{2} \sin 40 t\right) .\) The initial conditions \(q(0)=5\) and \(q^{\prime}(0)=0\) imply \(c_{1}=5\) and \(c_{2}=5 / 2 .\) Thus \\[q(t)=e^{-20 t}\left(5 \cos 40 t+\frac{5}{2} \sin 40 t\right)=\sqrt{25+25 / 4} e^{-20 t} \sin (40 t+1.1071)\\] and \(q(0.01) \approx 4.5676\) coulombs. The charge is zero for the first time when \(40 t+1.1071=\pi\) or \(t \approx\) 0.0509 second.

We are given that \(y_{1}=x^{2}\) is a solution of \(x^{4} y^{\prime \prime}+x^{3} y^{\prime}-4 x^{2} y=0 .\) To find a second solution we use reduction of order. Let \(y=x^{2} u(x)\). Then the product rule gives $$y^{\prime}=x^{2} u^{\prime}+2 x u \text { and } y^{\prime \prime}=x^{2} u^{\prime \prime}+4 x u^{\prime}+2 u,$$ so $$x^{4} y^{\prime \prime}+x^{3} y^{\prime}-4 x^{2} y=x^{5}\left(x u^{\prime \prime}+5 u^{\prime}\right)=0.$$ Letting \(w=u^{\prime},\) this becomes \(x w^{\prime}+5 w=0 .\) Separating variables and integrating we have $$\frac{d w}{w}=-\frac{5}{x} d x \quad \text { and } \quad \ln |w|=-5 \ln x+c.$$ Thus, \(w=x^{-5}\) and \(u=-\frac{1}{4} x^{-4} .\) A second solution is then \(y_{2}=x^{2} x^{-4}=1 / x^{2},\) and the general solution of the homogeneous differential equation is \(y_{c}=c_{1} x^{2}+c_{2} / x^{2}\). To find a particular solution, \(y_{p},\) we use variation of parameters. The Wronskian is $$W=\left|\begin{array}{cc}x^{2} & 1 / x^{2} \\\2 x & -2 / x^{3} \end{array}\right|=-\frac{4}{x}.$$ Identifying \(f(x)=1 / x^{4}\) we obtain \(u_{1}^{\prime}=\frac{1}{4} x^{-5}\) and \(u_{2}^{\prime}=-\frac{1}{4} x^{-1} .\) Then \(u_{1}=-\frac{1}{16} x^{-4}\) and \(u_{2}=-\frac{1}{4} \ln x,\) so $$y_{p}=-\frac{1}{16} x^{-4} x^{2}-\frac{1}{4}(\ln x) x^{-2}=-\frac{1}{16} x^{-2}-\frac{1}{4} x^{-2} \ln x.$$ The general solution is $$y=c_{1} x^{2}+\frac{c_{2}}{x^{2}}-\frac{1}{16 x^{2}}-\frac{1}{4 x^{2}} \ln x.$$

The auxiliary equation is \(m^{2}=0\) so that \(u(r)=c_{1}+c_{2} \ln r .\) The boundary conditions \(u(a)=u_{0}\) and \(u(b)=u_{1}\) yield the system \(c_{1}+c_{2} \ln a=u_{0}, c_{1}+c_{2} \ln b=u_{1} .\) Solving gives \\[ c_{1}=\frac{u_{1} \ln a-u_{0} \ln b}{\ln (a / b)} \quad \text { and } \quad c_{2}=\frac{u_{0}-u_{1}}{\ln (a / b)} \\]. Thus \\[ u(r)=\frac{u_{1} \ln a-u_{0} \ln b}{\ln (a / b)}+\frac{u_{0}-u_{1}}{\ln (a / b)} \ln r=\frac{u_{0} \ln (r / b)-u_{1} \ln (r / a)}{\ln (a / b)} \\].

Identifying \(P(x)=2(1+x) /\left(1-2 x-x^{2}\right)\) we have $$y_{2}=(x+1) \int \frac{e^{-\int 2(1+x) d x /\left(1-2 x-x^{2}\right)}}{(x+1)^{2}} d x=(x+1) \int \frac{e^{\ln \left(1-2 x-x^{2}\right)}}{(x+1)^{2}} d x$$ $$=(x+1) \int \frac{1-2 x-x^{2}}{(x+1)^{2}} d x=(x+1) \int\left[\frac{2}{(x+1)^{2}}-1\right] d x$$ $$=(x+1)\left[-\frac{2}{x+1}-x\right]=-2-x^{2}-x$$ A second solution is \(y_{2}=x^{2}+x+2\)

(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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