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Consider the differential equation \(y^{\prime \prime}+a y^{\prime}+b y=0\), where \(a\) and \(b\) are positive real constants. Show that \(\lim _{t \rightarrow \infty} y(t)=0\) for every solution of this equation.

Short Answer

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Question: Show that every solution of the given second-order linear homogeneous differential equation with constant coefficients converges to 0 as time goes to infinity. Given differential equation: \(y^{\prime \prime} + ay^{\prime} + by = 0\), where \(a\) and \(b\) are positive constants.

Step by step solution

01

Write down the given differential equation

The given differential equation is: $$y^{\prime \prime} + a y^{\prime} + b y = 0$$
02

Find the characteristic equation

To find the general solution of this linear homogeneous differential equation, we first need to find the characteristic equation. Replace \(y^{\prime \prime}\) with \(m^2\), \(y^{\prime}\) with \(m\), and \(y\) with 1: $$m^2 + am + b = 0$$
03

Find the roots of the characteristic equation

The roots of the characteristic equation determine the behavior of the general solution. Using the quadratic formula, the two roots are given by: $$m_{1,2} = \frac{-a \pm \sqrt{a^2 - 4b}}{2}$$
04

Identify different cases for the roots

Depending on the discriminant (\(a^2 - 4b\)), we will have different cases for the roots: 1. When \(a^2 - 4b > 0\), the roots are real and distinct. 2. When \(a^2 - 4b = 0\), the roots are real and the same. 3. When \(a^2 - 4b < 0\), the roots are complex.
05

Find the general solution for different cases

For each case, we can find the general solution of the differential equation: 1. Real and distinct roots: $$y(t) = c_1 e^{m_1 t} + c_2 e^{m_2 t}$$ 2. Real and the same roots: $$y(t) = (c_1 + c_2 t)e^{m_1 t}$$ 3. Complex roots: $$y(t) = e^{rt}(c_1 \cos(\omega t) + c_2 \sin(\omega t))$$, where \(m_1 = r + i\omega\) and \(m_2 = r - i\omega\).
06

Investigate the behavior of the general solution as \(t \rightarrow \infty\)

We will now analyze each case separately. 1. Real and distinct roots: Because \(a\) and \(b\) are positive, both \(m_1\) and \(m_2\) are negative. As \(t\) goes to infinity, both terms approach 0: $$\lim_{t \rightarrow \infty}y(t) = \lim_{t \rightarrow \infty}(c_1 e^{m_1 t} + c_2 e^{m_2 t}) = 0$$ 2. Real and the same roots: Similarly, as \(a\) and \(b\) are positive, \(m_1\) is also negative. Both terms of the solution approach 0 as \(t\) goes to infinity: $$\lim_{t \rightarrow \infty}y(t) = \lim_{t \rightarrow \infty}((c_1 + c_2 t) e^{m_1 t}) = 0$$ 3. Complex roots: In this case, \(r\) is negative, so the \(e^{rt}\) term approaches 0 as \(t \rightarrow \infty\). The oscillatory part \(\cos(\omega t)\) and \(\sin(\omega t)\) are bounded between -1 and 1. Thus, $$\lim_{t \rightarrow \infty}y(t) = \lim_{t \rightarrow \infty}(e^{rt}(c_1 \cos(\omega t) + c_2 \sin(\omega t))) = 0$$ In each of the three cases, the limit of the general solution as \(t \rightarrow \infty\) is 0, which completes the proof.

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Key Concepts

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

Characteristic Equation
The characteristic equation is a pivotal concept in solving linear homogeneous differential equations. It's essentially a polynomial whose roots determine the general solution to the differential equation. For instance, consider a second-order homogeneous differential equation of the form
\[y'' + ay' + by = 0,\]
where a and b are constants. To solve this, we introduce a new variable, typically denoted m, which represents the potential values for the rate of growth or decay in our solutions. The characteristic equation is then formed by substituting m for each derivative of y, leading to: \[m^2 + am + b = 0.\]
The roots of this quadratic equation, found using the quadratic formula, guide the form of the solution to the original differential equation.
Homogeneous Differential Equation
A homogeneous differential equation is one in which every term is a function of the variable and its derivatives, and it is set to zero. This means all terms can be expressed in terms of the dependent variable and its derivatives. For the homogeneous second-order linear differential equation given by \[y'' + ay' + by = 0,\]
the homogeneous nature means that we are dealing with a system that, when undisturbed by external forces, returns to rest – symbolized by the zero on the right side of the equation. The focus is on the natural behavior of the system. The general solutions are derived from the roots of the characteristic equation and represent the fundamental modes of the system's free response. Because the roots play such a critical role, they are responsible for deciding whether the solution will display exponential growth, decay, or oscillation.
Complex Roots of Differential Equations
When the characteristic equation yields complex roots, it indicates oscillatory behavior in the solutions of the differential equation. If the characteristic equation is \[m^2 + am + b = 0\]
and the discriminant a^2 - 4b is less than zero, we have complex roots of the form \[m = r \text{pm} i\text{omega},\]
where r is the real part and omega is the imaginary part (related to the frequency of oscillation). In this case, the general solution involves trigonometric functions and is represented as \[y(t) = e^{rt}(c_1 \text{cos}(\text{omega} t) + c_2 \text{sin}(\text{omega} t)).\]
The real part r determines the exponential growth or decay, while the sine and cosine terms represent the oscillatory component. If the real part is negative, as in our problem where a and b are positive, the exponential component decreases over time, leading to the conclusion that \[\text{lim}_{{t \text{right arrow} \text{infinity}} y(t) = 0,\]
regardless of the oscillations. This ensures that the solution tends toward zero as time goes to infinity, which can be critical information when predicting long-term behavior of a system described by such an equation.

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Most popular questions from this chapter

For each differential equation, (a) Find the complementary solution. (b) Formulate the appropriate form for the particular solution suggested by the method of undetermined coefficients. You need not evaluate the undetermined coefficients. $$ y^{\prime \prime \prime}-4 y^{\prime \prime}+4 y^{\prime}=t^{3}+4 t^{2} e^{2 t} $$

In each exercise, you are given the general solution of $$ y^{(4)}+a_{3} y^{\prime \prime \prime}+a_{2} y^{\prime \prime}+a_{1} y^{\prime}+a_{0} y=0, $$ where \(a_{3}, a_{2}, a_{1}\), and \(a_{0}\) are real constants. Use the general solution to determine the constants \(a_{3}, a_{2}, a_{1}\), and \(a_{0}\). [Hint: Construct the characteristic equation from the given general solution.] $$ y(t)=c_{1} e^{t}+c_{2} t e^{t}+c_{3} e^{-t}+c_{4} t e^{-t} $$

In each exercise, (a) Find the general solution of the differential equation. (b) If initial conditions are specified, solve the initial value problem. $$ 2 y^{\prime \prime \prime}-y^{\prime \prime}=0 $$

(a) Obtain the general solution of the differential equation. (b) Impose the initial conditions to obtain the unique solution of the initial value problem. (c) Describe the behavior of the solution as \(t \rightarrow-\infty\) and \(t \rightarrow \infty\). In each case, does \(y(t)\) approach \(-\infty,+\infty\), or a finite limit? $$ 25 y^{\prime \prime}+20 y^{\prime}+4 y=0, \quad y(5)=4 e^{-2}, \quad y^{\prime}(5)=-\frac{3}{5} e^{-2} $$

(a) Verify that the given function, \(y_{P}(t)\), is a particular solution of the differential equation. (b) Determine the complementary solution, \(y_{C}(t)\). (c) Form the general solution and impose the initial conditions to obtain the unique solution of the initial value problem. $$y^{\prime \prime}-2 y^{\prime}-3 y=-9 t-3, \quad y(0)=1, \quad y^{\prime}(0)=3, \quad y_{p}(t)=3 t-1$$

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