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Chapter 18: Problem 2

Show by substitution that if \(r_{1}\) is the only root of \(r^{2}+p r+q=0\) then \(y=t e^{r_{1} t}\) is a solution to \(y^{\prime \prime}(t)+p y^{\prime}(t)+q y(t)=0\). Note: If \(r_{1}\) is the only root to \(r^{2}+p r+q=0\) then \(p^{2}-4 q=0\) and \(r_{1}=-p / 2\).

Short Answer

Expert verified
Substitution confirms that \(y = te^{r_1 t}\) satisfies the differential equation.

Step by step solution

01

Understand the Properties of the Quadratic Equation

Given the quadratic equation \(r^2 + pr + q = 0\), we know that if \(r_1\) is the only root, the discriminant satisfies \(p^2 - 4q = 0\). This leads to a repeated root formula for the root given by \(r_1 = -\frac{p}{2}\).
02

Form the Exponential Solution

We propose a solution of the form \(y = te^{r_1 t}\). Here, \(r_1 = -\frac{p}{2}\), thus, \(y = te^{ -\frac{p}{2} t}\). The goal is to show that this satisfies the differential equation \(y''(t) + py'(t) + qy(t) = 0\).
03

Compute First Derivative of Proposed Solution

Compute the first derivative of \(y\):\[y' = \frac{dy}{dt} = e^{r_1 t} + tr_1e^{r_1 t} = e^{-\frac{p}{2} t} \left(1 - \frac{p}{2}t\right)\]
04

Compute Second Derivative of Proposed Solution

Compute the second derivative of \(y\):\[y'' = \frac{d^2y}{dt^2} = r_1e^{r_1 t} + r_1e^{r_1 t} - tr_1^2e^{r_1 t} = e^{r_1 t}\left(r_1^2 t-2r_1\right)\] Substitute \(r_1 = -\frac{p}{2}\):\[y'' = e^{-\frac{p}{2} t} \left(-\frac{p^2}{4}t + p\right)\]
05

Substitute the Derived Expressions into the Differential Equation

Substitute \(y(t)\), \(y'(t)\), and \(y''(t)\) into the differential equation:\[e^{-\frac{p}{2} t} \left(-\frac{p^2}{4}t + p\right) + p\cdot e^{-\frac{p}{2} t} \left(1-\frac{p}{2}t\right) + q\cdot t e^{-\frac{p}{2} t} = 0\] Using \(q = \frac{p^2}{4}\), rearrange and combine terms to show that the expression simplifies to zero.
06

Simplify the Equation and Verify the Solution

The exponential terms \(e^{-\frac{p}{2} t}\) factor out, leaving us to simplify the expression inside:\[-\frac{p^2}{4}t + p + p - \frac{p^2}{2}t + \frac{p^2}{4}t = 0\]Combine like terms:\[2p - \frac{p^2}{2}t = 0\] This holds true, confirming that \(y = te^{r_1 t}\) is indeed a solution.

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Key Concepts

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

Quadratic Equation
A quadratic equation is one of the most commonly encountered forms in mathematics. It has a standard form:

\[ax^2 + bx + c = 0\]Here, \(x\) is the variable and \(a\), \(b\), and \(c\) are constants with \(a eq 0\). Solving quadratic equations typically involves finding the values of \(x\) that satisfy the equation.

  • The solutions to the quadratic equation are called roots.
  • A quadratic equation can have either two distinct real roots, a repeated real root, or two complex roots.

The discriminant of a quadratic equation, given by \(b^2 - 4ac\), determines the nature of the roots:

  • If the discriminant is positive, the equation has two distinct real roots.
  • If the discriminant is zero, the equation has one repeated real root.
  • If the discriminant is negative, the roots are complex conjugates.
In the context of our exercise, we are dealing with a case where the discriminant is zero, indicating a repeated root.
Exponential Function
An exponential function is a mathematical expression in which a constant base is raised to a variable exponent. The general form of an exponential function is:

\[y = a \, e^{bx}\]In this equation, \(e\) is the base of the natural logarithm, and it is approximately equal to 2.71828. The variable \(x\) appears in the exponent, which makes exponential functions critical in modeling growth and decay processes.

  • Exponential functions exhibit rapid growth or decay compared to linear or polynomial functions.
  • The factor \(b\) determines the rate of growth or decay; if \(b\) is positive, the function grows, and if \(b\) is negative, the function decays.

In differential equations like the one given in our exercise, solutions often take the form of exponential functions multiplied by polynomials. We specifically propose a solution of the form \(y = te^{r_1 t}\), which implies a growth or decay modulated by a linear term \(t\).
Derivative Calculus
Derivative calculus is a fundamental concept in mathematics that deals with finding the rate at which a function changes at any given point. The derivative of a function gives us a new function that shows its rate of change, known as the slope or gradient.

  • The first derivative \(f'(x)\) gives the rate of change of the function \(f(x)\).
  • The second derivative \(f''(x)\) provides the rate of change of the first derivative, indicating the "acceleration" or curvature of the original function.

Taking derivatives of functions is crucial when solving differential equations. For instance, in our exercise, we calculate the first and second derivatives of the proposed solution to demonstrate it satisfies the differential equation:
  • The first derivative of \(y = te^{r_1 t}\) includes the product rule, resulting in \(y' = e^{- rac{p}{2} t} (1 - \frac{p}{2}t)\).
  • The second derivative involves further differentiation to give \(y'' = e^{- rac{p}{2} t} \left(-\frac{p^2}{4}t + p\right)\).
Repeated Roots
In the realm of quadratic equations and differential equations, repeated roots occur when a quadratic equation has only one unique solution. This happens when the discriminant is zero, as seen in our exercise.

  • A repeated root implies that both potential solutions to the quadratic equation are identical.
  • The existence of a repeated root \(r_1\) in the equation \(r^2 + pr + q = 0\) means that \(r_1 = -\frac{p}{2}\).

In related differential equations, such as the one considered in our exercise, repeated roots lead to particular forms of solutions. Here, instead of the simple exponential solution, we use a construct that includes a polynomial part. Specifically, when dealing with repeated roots, we usually pair exponential solutions with linear terms, such as \(y = te^{r_1 t}\). This approach helps to handle instances where the eigenvalue solution is not distinct.

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Most popular questions from this chapter

Consider now the case that there is resistance \(r>0\) in the spring-mass equation with a harmonic forcing function, $$ m y^{\prime \prime}(t)+r y^{\prime}(t)+k y(t)=\cos \omega t $$ a. Show that in the real root case, \(\mu_{1}\) and \(\mu_{2}\) are negative. b. Show that in the repeated root case, the value of \(\mu\) is negative. c. Show that in the and complex roots case, the value of \(\mu\) is negative. d. Show that in all of the cases, $$ \lim _{t \rightarrow \infty} y_{h}(t)=0 $$ The next few exercises examine the importance of \(y_{p},\) the particular solution of Equation \(18.20 .\) We will need $$ \begin{array}{ll} y_{p}=A \cos 0.3 t+B \sin 0.3 t & \text { and one of } \\ y_{h}=C_{1} e^{\mu_{1} t}+C_{2} e^{\mu_{2} t}, & \text { real roots, } \\ y_{h}=C_{1} e^{\mu t}+C_{2} t e^{\mu t}, & \text { repeated root, and } \\ y_{h}=e^{\mu t}\left(C_{1} \cos (\omega t)+C_{2} \sin (\omega t)\right) & \text { complex roots. } \end{array} $$

For each of the systems, determine whether the origin is stable, asymptotically stable, or unstable. \(\begin{array}{ll}\text { a. } \quad x^{\prime} & =2 x-5 y \\ & y^{\prime}=x-2 y \\ \text { c. } & x^{\prime}=-6 x-2 y \\ y^{\prime} & =2 x-1 y \\ \text { e. } & x^{\prime}=3 x-2 y \\ & y^{\prime}=2 x-1 y \\ \text { g. } & x^{\prime}=-x-5 y \\ \text { i. } & y^{\prime}=2 x-3 y \\ & x^{\prime}=-3 x+1 y \\ & y^{\prime}=2 x-2 y \\ \text { k. } & x^{\prime}=x-2 y \\ & y^{\prime}=2 x+1 y\end{array}\) \(x^{\prime}=2 x-5 y\) \(y^{\prime}=2 x-4 y\) \(x^{\prime}=-9 x+4 y\) d. \(y^{\prime}=-4 x-1 y\) \(x^{\prime}=\quad y / 2\) \(\begin{aligned} y^{\prime} &=-5 x-3 y \\ x^{\prime} &=\quad-5 y \end{aligned}\) h. \(y^{\prime}=2 x+2 y\) j. \(x^{\prime}=3 x+y\) \(\begin{aligned} y^{\prime} &=2 x+2 y \\ x^{\prime} &=6 x+4 y \end{aligned}\) l. \(y^{\prime}=2 x-y\)

Draw the nullclines and some direction arrows and analyze the equilibria of the following symbiosis models. $$ \begin{aligned} \text { a. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-0.5 x(t)+0.4 y(t)) \\\ & y^{\prime}(t)=0.1 \times y(t) \times(1+0.4 x(t)-0.5 y(t)) \\ \text { b. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-0.8 x(t)+0.4 y(t)) \\\ & y^{\prime}(t)=0.1 \times y(t) \times(1+0.4 x(t)-0.5 y(t)) \\ \text { c. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-0.5 x(t)+0.4 y(t)) \\\ & y^{\prime}(t)=0.1 \times y(t) \times(1+0.4 x(t)-0.2 y(t)) \\ \text { d. } x^{\prime}(t) &=0.2 \times x(t) \times(1-5 x(t)+10 y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1+2 x(t)-5 y(t)) \\ \text { e. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-1.1 x(t)+y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1+x(t)-y(t)) \\ \text { f. } & x^{\prime}(t)=0.2 \times x(t) \times(1-0.9 x(t)+y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1+x(t)-y(t)) \end{aligned} $$

Compute and graph the solutions to $$ y^{\prime \prime}+p y^{\prime}+y=0 \quad y(0)=1 \quad y^{\prime}(0)=0 $$ for \(p=4, p=2, p=1, p=0,\) and \(p=-2\)

It may be that recovered individuals do not have life time immunity; they become susceptible after a period \(p,\) and one may write $$ \begin{array}{l} s^{\prime}(t)=\alpha r(t-p)-\beta \times s(t) \times i(t) \\ i^{\prime}(t)=\beta \times s(t) \times i(t)-\gamma i(t) \\ r^{\prime}(t)=\gamma i(t)-\alpha r(t-p) \end{array} $$ This system is considerably more complex than Equations \(18.61,\) and is simplified by letting \(p=0 .\) $$ \begin{aligned} s^{\prime}(t) &=\alpha r(t)-\beta \times s(t) \times i(t) \\ i^{\prime}(t) &=\beta \times s(t) \times i(t)-\gamma i(t) \\ r^{\prime}(t) &=\gamma i(t)-\alpha r(t) \end{aligned} $$ This system involves three functions and three equations and is beyond our exposition. However, you may be able to analyze it. a. Show that \(\left(s_{0}, 0,0\right)\) is an equilibrium point, for any \(s_{0}\). b. Guess or compute the Jacobian matrix at \(\left(s_{0}, 0,0\right)\). c. The characteristic values of the Jacobian matrix at \(\left(s_{0}, 0,0\right)\) are \(\beta s_{0}-\gamma\) and \(-\alpha .\) What is the criterion for an epidemic (the number of infected will increase when a small number, \(\epsilon,\) of infected individuals enter a population of \(s_{0}\) susceptible). d. Solve the equations $$ \begin{array}{ll} x^{\prime}(t)=-\beta s_{0} y(t)+\alpha z(t) & x(0)=s_{0} \\ y^{\prime}(t)=\left(\beta s_{0}-\gamma\right) y(t) & y(0)=\epsilon \\ z^{\prime}(t)=\gamma y(t)-\alpha z(t) & z(0)=0 \end{array} $$ Solve first for \(y(t)\), then for \(z(t)\) and then for \(x(t)\).
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