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Refer to the equation $$x^{\prime \prime}+b x^{\prime}+c x=0 .$$ Suppose \(b<0\) and \(c>0 .\) For \(x(0)\) and \(x^{\prime}(0)\) not both zero, is it possible that \(\lim _{t \rightarrow \infty}|x(t)|=L\), where \(L\) is finite? Explain.

Short Answer

Expert verified
No, it is not possible for the limit of |x(t)| as t approaches infinity to be a finite value L other than zero.

Step by step solution

01

Analyzing the differential equation

The given equation is a second order linear homogeneous differential equation. The general solution to this equation can be found by finding the roots of the characteristic equation, \[m^2+bm+c=0.\] By using the quadratic formula, the roots become \[m_{1,2}=(-b \pm \sqrt{b^2 - 4c}) / 2.\] Depending on the discriminant \(b^2 - 4c\), there are three cases: two real distinct roots if \(b^2 - 4c > 0\), two identical real roots if \(b^2 - 4c = 0\), or complex roots if \(b^2 - 4c < 0\).
02

Determining the type of roots

The exercise mentions \(b < 0\) and \(c > 0\). This tells us that the discriminant \(b^2 - 4c\) will always be negative, because b squared (a positive value) is less than 4 times c (also a positive value), implying that the roots of the characteristic equation will be complex.
03

Evaluating the general solution

In the case of complex roots, the general solution of the differential equation is \[x(t) = e^{rt}(A\cos(\omega t) + B\sin(\omega t)).\] Here, \(r\) and \(\omega\) are respectively the real and the imaginary part of the complex roots. We know that \(r = -b/2 < 0\) from the calculation of the roots. Thus, the exponential part will decay as t approaches to infinity.
04

Conclude the solution

Since \(r = -b/2 < 0\), as \(t \rightarrow \infty\), \(e^{rt} \rightarrow 0\), while the \((A\cos(\omega t) + B\sin(\omega t))\) part oscillates between -1 and 1. Therefore, \[ \lim_{t \rightarrow \infty}|x(t)|=0 \] regardless of the values of initial conditions \(x(0)\) and \(x'(0)\) that are not both zero. It is thus concluded that it is not possible for the limit of |x(t)| as t approaches infinity to be a finite value L other than 0.

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Key Concepts

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

Characteristic Equation
When working with second order linear homogeneous differential equations, a key concept is the characteristic equation. For the equation \[ x'' + bx' + c x = 0, \] the characteristic equation is derived by assuming a solution of the form \( x(t) = e^{mt} \). Substituting into the differential equation results in a quadratic equation in \( m \):\[ m^2 + bm + c = 0. \]Solving this characteristic equation is crucial since the roots, \( m_1 \) and \( m_2 \), determine the behavior of the solutions of the differential equation.
  • If \( b^2 - 4c > 0 \), we get two distinct real roots.
  • If \( b^2 - 4c = 0 \), we get a repeated real root.
  • If \( b^2 - 4c < 0 \), we encounter complex roots.
These roots are pivotal in shaping the form of the general solution to the differential equation. Understanding how to form and solve the characteristic equation is a foundational skill in solving these equations.
Complex Roots
In the context of linear differential equations, when the discriminant \( b^2 - 4c \) is negative, the roots of the characteristic equation are complex. This happens under certain conditions of the coefficients, as seen in the exercise where \( b < 0 \) and \( c > 0 \), making \( b^2 - 4c \) always negative.The complex roots will be in the form:\[ m_1, m_2 = (-b \pm i \sqrt{4c - b^2}) / 2. \]This means each root has a real part, \( r = -b/2 \), and an imaginary part, \( \omega = \sqrt{4c - b^2}/2 \). The general solution involving complex roots is:\[ x(t) = e^{rt}(A\cos(\omega t) + B\sin(\omega t)), \]where:
  • \( e^{rt} \) is an exponential term that controls the amplitude of the solution.
  • \(\cos(\omega t)\) and \(\sin(\omega t)\) represent the oscillatory nature of the solution due to the imaginary part \( \omega \).
Since \( r < 0 \), \( e^{rt} \) approaches zero as \( t \) tends to infinity, leading the entire solution towards zero, illustrating the decaying oscillation.
Homogeneous Differential Equation
A homogeneous differential equation is one where all terms are made up of the function \( x(t) \) and its derivatives, and there are no independent terms. In this problem,\[ x'' + bx' + c x = 0, \]is such a homogeneous differential equation. It is called homogeneous because the right-hand side is zero. This equation means that out of all solutions, we are particularly interested in those whose behavior depends solely on the relationship and interaction between \( x(t) \), its derivatives, and the coefficients in the equation.Solving a homogeneous differential equation involves finding the characteristic equation and solving for its roots, leading to the general solution:\[ x(t) = e^{rt}(A\cos(\omega t) + B\sin(\omega t)), \]as seen from the exercise, especially when roots are complex.Key characteristics of homogeneous equations include:
  • The solution set forms a vector space.
  • Superposition principle applies — sums of solutions are also solutions.
Understanding these allows us to find the full spectrum of solutions that satisfy both the differential equation and given initial conditions.

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Most popular questions from this chapter

Solve the following. (a) \(\frac{d x}{d t}=6-2 x\). Do this using substitution in two ways. i. Factor out a \(-2\) from the right-hand side and let \(u=x-3\). Then solve. ii. Let \(v=6-2 x .\) Express \(\frac{d x}{d t}\) in terms of \(v\) and then convert the equation \(\frac{d x}{d t}=6-2 x\) to an equation in \(v\) and \(\frac{d v}{d t}\) and solve. (b) \(\frac{d x}{d t}=3 x-7\) (c) \(\frac{d y}{d t}=k y+B\)

Each function below is a solution to one of the second orderdifferential equations listed. To each function match the appropriate differential equation. \(C_{1}\) and \(C_{2}\) are constants. Differential Equations I. \(\frac{d^{2} x}{d t^{2}}-9 x=0\) II. \(\frac{d^{2} x}{d t^{2}}+9 x=0\) III. \(\frac{d^{2} x}{d t^{2}}=3 x\) Solution Functions (a) \(x(t)=5 e^{3 t}\) (b) \(x(t)=-2 e^{\sqrt{3} t}\) (c) \(x(t)=7 \sin 3 t\) (d) \(x(t)=C_{1} \sin 3 t+C_{2} \cos 3 t\) (e) \(x(t)=C_{1} e^{\sqrt{3} t}+C_{2} e^{-\sqrt{3} t}\)

\(P(t)=C e^{k t}+\frac{E}{k}\), where \(C\) is a constant, is the general solution to the differential equation \(\frac{d P}{d t}=k P-E .\) Below is the slope eld for \(\frac{d P}{d t}=2 P-6\). (a) i. Find the particular solution that corresponds to the initial condition \(P(0)=2\). ii. Sketch the solution curve through \((0,2)\). (b) i. Find the particular solution that corresponds to the initial condition \(P(0)=3\). ii. Sketch the solution curve through \((0,3)\). (c) i. Find the particular solution that corresponds to the initial condition \(P(0)=4\). ii. Sketch the solution curve through \((0,4)\).

Money is deposited in a bank account with a nominal annual interest rate of \(4 \%\) compounded continuously. Let \(M=M(t)\) be the amount of money in the account at time \(t\). (a) Write a differential equation whose solution is \(M(t)\). Assume there are no additional deposits and no withdrawals. (b) Suppose money is being added to the account continuously at a rate of \(\$ 1000\) per year and no withdrawals are made. Write a differential equation whose solution is \(M(t)\)

A very large container of juice contains four gallons of apple juice and one gallon of cranberry juice. Cranberry-apple juice \((60 \%\) apple, \(40 \%\) cranberry) is entering the container at a rate of three gallons per hour. The well-stirred mixture is leaving the container at three gallons per hour. (a) Write a differential equation whose solution is \(C(t)\), the number of gallons of cranberry juice in the container at time \(t .\) Solve, using the initial condition. (b) Write a differential equation whose solution is \(A(t)\), the number of gallons of apple juice in the container at time \(t .\) Solve the equation.

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