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Chapter 9: Problem 9

Solve. \(y^{\prime \prime}-9 y=0\)

Short Answer

Expert verified
The general solution is \(y(t) = C_1e^{3t} + C_2e^{-3t}\).

Step by step solution

01

Recognize the type of differential equation

This is a second-order linear homogeneous differential equation of the form \(y'' - 9y = 0\).
02

Write characteristic equation

To solve this, we first write down its characteristic equation. For a differential equation of the form \(ay'' + by' + cy = 0\), the characteristic equation is given by \(ar^2 + br + c = 0\).Given \(y'' - 9y = 0\), the characteristic equation is: \(r^2 - 9 = 0\).
03

Solve the characteristic equation

Solve \(r^2 - 9 = 0\) using factoring or the quadratic formula.Factoring it, we get: \((r - 3)(r + 3) = 0\).So, the roots are \(r = 3\) and \(r = -3\).
04

Write the general solution

Given the roots \(r = 3\) and \(r = -3\), the general solution to the differential equation is:\(y(t) = C_1e^{3t} + C_2e^{-3t}\), where \(C_1\) and \(C_2\) are constants that can be determined by initial conditions.

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Key Concepts

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

Characteristic Equation
When solving second-order linear homogeneous differential equations, an essential step is formulating the characteristic equation. The standard form of such a differential equation is expressed as:
To determine the characteristic equation from this, substitute each derivative with polynomials involving the variable r. This leads us to:
  • ar^2 + br + c = 0
For the given problem, the equation is:
  • y'' - 9y = 0
By identifying that a = 1, b = 0, and c = -9, we obtain the characteristic equation:
  • r^2 - 9 = 0
The characteristic equation is crucial because solving it helps find the roots that are used to build the general solution for the differential equation.
Homogeneous Differential Equation
A homogeneous differential equation has terms that all involve the function y or its derivatives. There's no standalone function of the independent variable (such as t in this case). The given differential equation:
  • y'' - 9y = 0
doesn't include any terms that are functions of t alone, thus making it homogeneous. The solution to a homogeneous differential equation typically involves finding a general solution from the roots of the characteristic equation. These roots form the basis of the solution in exponential terms of e. Here, since we have the characteristic equation:
  • r^2 - 9 = 0
Solving for r yields roots that are fundamental to constructing the general solution of the equation.
Roots of Polynomial Equations
Finding the roots of polynomial equations is a central part of solving differential equations. For our characteristic equation
  • r^2 - 9 = 0
we solve by factoring as follows:
  • (r - 3)(r + 3) = 0
The solutions or roots for this equation are r = 3 and r = -3. These roots might be real and distinct, real and repeated, or complex, each leading to different forms of the general solution. For distinct real roots, as seen here, the general solution combines exponential terms using these roots:
  • y(t) = C_1e^{3t} + C_2e^{-3t}
where C_1 and C_2 are constants. These constants can later be determined if initial conditions are provided.

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Most popular questions from this chapter

Find the equilibrium points and assess the stability of each. \(x^{\prime}=x^{2}+y^{2}-25, y^{\prime}=x+y-7\)

Find the equilibrium points and assess the stability of each. \(x^{\prime}=x-e^{y}, y^{\prime}=2 \ln x+y-6\)

Suppose a reactant has two isomeric forms \(A\) and \(B\). The reactant initially has concentration \(q\) and is entirely in form \(A\), but may freely convert from form \(A\) to form \(B\) and back again. Also, once in form \(A\), the reactant may produce two products \(C\) and \(D\). The concentration \(F(t)\) of product \(C\) after \(t\) minutes satisfies the second-order differential equation $$ F^{\prime \prime}+(a+b+c+d) F^{\prime}+b(c+d) F=b c q $$ where the constants \(a, b, c\), and \(d\) describe the rates that \(A, B, C\), and \(D\) are produced. Also, \(F(0)=0\) and \(F^{\prime}(0)=c q\) initially. Suppose that \(a=0, b=1, c=0.05, d=0\), and \(q=1 .\) Find \(F(t)\)

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