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Chapter 3: Problem 21

find the fundamental set of solutions specified by Theorem 3.2.5 for the given differential equation and initial point. $$ y^{\prime \prime}+y^{\prime}-2 y=0, \quad t_{0}=0 $$

Short Answer

Expert verified
The fundamental set of solutions for the given second-order differential equation and initial point is: $$ y(t) = c_1e^{-2t} + c_2e^t $$

Step by step solution

01

Write down the characteristic equation for the given differential equation

The given differential equation is: $$ y^{\prime\prime}+y^{\prime}-2y=0 $$ The corresponding characteristic equation is: $$ r^2 + r - 2 = 0 $$
02

Solve the characteristic equation

To solve the quadratic equation: $$ r^2 + r - 2 = 0 $$ We can factor it as: $$ (r + 2)(r - 1) = 0 $$ Hence, the two roots of the characteristic equation are: \(r_1 = -2\) and \(r_2 = 1\).
03

Write down the fundamental set of solutions using the roots of the characteristic equation

Since we have two distinct real roots, the fundamental set of solutions for the given differential equation is: $$ y(t) = c_1e^{-2t} + c_2e^t, $$ where \(c_1\) and \(c_2\) are constants.
04

Use the initial point to find the particular solution

The given initial point is \(t_0 = 0\). To find the particular solution, substitute the initial point into the general solution: $$ y(0) = c_1e^{0} + c_2e^{0} $$ Since \(e^0 = 1\), the equation becomes: $$ y(0) = c_1 + c_2 $$ However, we do not have the value for \(y(0)\), so we cannot directly determine the values for \(c_1\) and \(c_2\). Therefore, we'll leave the particular solution as: $$ y(t) = c_1e^{-2t} + c_2e^t $$
05

Provide the fundamental set of solutions as specified by Theorem 3.2.5

The fundamental set of solutions specified by Theorem 3.2.5 for the given differential equation and initial point is: $$ y(t) = c_1e^{-2t} + c_2e^t $$

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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 key concept when solving linear homogeneous differential equations with constant coefficients. It is derived from the differential equation by replacing the derivatives of y with powers of a hypothetical solution r. For instance, if you have an equation like y'' + y' - 2y = 0, you create its characteristic equation by considering y(t) = e^{rt}, which leads to the algebraic equation r^2 + r - 2 = 0.

Solving this quadratic equation produces the roots, which are the essential elements for constructing the general solution of the differential equation. If the roots are real and distinct, the solution consists of exponentials raised to the power of these roots, which reflects different growth or decay rates in the solution. However, if the roots are complex or repeated, the form of the solution changes to accommodate terms involving sines, cosines, or polynomial factors.
Initial Value Problem
An initial value problem (IVP) involves not only a differential equation but also an initial condition that specifies the value of the unknown function (and possibly its derivatives) at a given point, typically denoted as t_0. This additional information allows us to find a specific solution, also known as the particular solution, from the general solution to the differential equation.

With the example at hand, the initial value is not provided; thus, we cannot determine the constants c_1 and c_2 without additional information. Specifically, you would need the value of y (and possibly y') at t_0 to find a unique solution. When dealing with initial value problems, always ensure that all necessary initial conditions are provided to find a complete, specific solution to the problem.
Fundamental Set of Solutions
The term fundamental set of solutions refers to a collection of independent solutions to a homogeneous linear differential equation that spans the solution space. In other words, any solution to the differential equation can be written as a linear combination of these fundamental solutions.

For the provided equation, the roots r_1 = -2 and r_2 = 1 lead to two distinct solutions e^{-2t} and e^t, respectively. Hence, our fundamental set is composed of these two exponential functions. By 'independent,' we mean that no solution in this set can be written as a scalar multiple or a combination of the others. This independence is crucial since it guarantees that the set provides the capability to generate the entirety of the solution space for the given differential equation.

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Most popular questions from this chapter

(a) Show that the phase of the forced response of Eq. ( 1) satisfies tan \(\delta=\gamma \omega / m\left(\omega_{0}^{2}-\omega^{2}\right)\) (b) Plot the phase \(\delta\) as a function of the forcing frequency \(\omega\) for the forced response of \(u^{\prime \prime}+0.125 u^{\prime}+u=3 \cos \omega t\)

A mass weighing 4 lb stretches a spring 1.5 in. The mass is displaced 2 in. in the positive direction from its equilibrium position and released with no initial velocity. Assuming that there is no damping and that the mass is acted on by an external force of \(2 \cos 3 t\) lb, formulate the initial value problem describing the motion of the mass.

If \(a, b,\) and \(c\) are positive constants, show that all solutions of \(a y^{\prime \prime}+b y^{\prime}+c y=0\) approach zero as \(t \rightarrow \infty\).

The differential equation $$ x y^{\prime \prime}-(x+N) y^{\prime}+N y=0 $$ where \(N\) is a nonnegative integer, has been discussed by several authors. 6 One reason it is interesting is that it has an exponential solution and a polynomial solution. (a) Verify that one solution is \(y_{1}(x)=e^{x}\). (b) Show that a second solution has the form \(y_{2}(x)=c e^{x} \int x^{N} e^{-x} d x\). Calculate \(y_{2 (x)\) for \(N=1\) and \(N=2 ;\) convince yourself that, with \(c=-1 / N !\) $$ y_{2}(x)=1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\cdots+\frac{x^{N}}{N !} $$ Note that \(y_{2}(x)\) is exactly the first \(N+1\) terms in the Taylor series about \(x=0\) for \(e^{x},\) that is, for \(y_{1}(x) .\)

Use the method of reduction of order to find a second solution of the given differential equation. \(x^{2} y^{\prime \prime}-(x-0.1875) y=0, \quad x>0 ; \quad y_{1}(x)=x^{1 / 4} e^{2 \sqrt{x}}\)
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