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Find the general solution to the given differential equation. $$9 z^{\prime \prime}-z=0$$

Short Answer

Expert verified
The general solution is \(z(t) = C_1 e^{\frac{1}{3} t} + C_2 e^{-\frac{1}{3} t}\).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is a second-order linear homogeneous differential equation with constant coefficients of the form \(az'' + bz' + cz = 0\). In this case, \(a = 9\), \(b = 0\), and \(c = -1\).
02

Write the Characteristic Equation

For a second-order linear homogeneous differential equation \(az'' + bz' + cz = 0\), the characteristic equation is obtained by replacing \(z''\) with \(r^2\), \(z'\) with \(r\), and \(z\) with a constant 1, giving us \(9r^2 - 1 = 0\).
03

Solve the Characteristic Equation

The characteristic equation \(9r^2 - 1 = 0\) simplifies to \(9r^2 = 1\). Divide by 9 to get \(r^2 = \frac{1}{9}\). Taking the square root of both sides, we find \(r = \pm\frac{1}{3}\).
04

Write the General Solution

Since both roots \(r_1 = \frac{1}{3}\) and \(r_2 = -\frac{1}{3}\) are real and distinct, the general solution to the differential equation is given by \(z(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}\). Substituting the values of \(r_1\) and \(r_2\), the solution becomes \(z(t) = C_1 e^{\frac{1}{3} t} + C_2 e^{-\frac{1}{3} t}\).

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Key Concepts

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

Second-Order Linear Differential Equation
A second-order linear differential equation involves a function and its derivatives up to the second order.
This type of equation is a vital concept in mathematics, especially because it models various physical phenomena such as oscillations and waves.
Let's break down the terminology:
  • "Second-order" refers to the highest derivative being the second derivative (\(z''\)). It indicates how the function evolves with second-degree changes.
  • "Linear" means the equation is linear in the unknown function and its derivatives. For example, the given differential equation, \(9 z^{\prime \prime}-z=0\), is linear. There are no powers or products of \(z, z', z''\).
The form of a second-order linear differential equation can be represented as:\[a z'' + b z' + c z = 0\]If the equation consists solely of constant coefficients, like in the example \(a = 9, b = 0, c = -1\), the analysis becomes straightforward. Constant coefficients mean that the coefficients of the derivatives do not vary with respect to the independent variable, which simplifies solving the equation.
Characteristic Equation
The characteristic equation is an auxiliary equation obtained by substituting a meaningful form of the solution into the differential equation.
This equation is crucial because it simplifies solving the differential equation.
For a given second-order linear differential equation:\[a z'' + b z' + c z = 0\]We replace the derivatives as follows:
  • \(z'' \rightarrow r^2\)
  • \(z' \rightarrow r\)
  • \(z \rightarrow 1\)
Doing so transforms our differential equation into the characteristic equation:\[9r^2 - 1 = 0\]Solving this equation keeps the process orderly and helps us find the roots, essential for determining the general solution.
In this context, the quadratic nature (\(r^2\)) reflects the second-order character, and the real roots (\(r = \pm\frac{1}{3}\)) hint at exponential growth or decay, leading us to the solution's structure.
Linear Homogeneous Differential Equation
A linear homogeneous differential equation is one where every term is either the unknown function or its derivatives.
Such an equation doesn't include terms independent of the unknown function (i.e. there are no standalone terms on the right side of the equation).
Considering our example:\[9 z^{\prime \prime}-z=0\]"Homogeneous" implies that if we substitute the zero function for the unknown function, the equation holds true.
This property tells us that the symmetries in the differential equation form solutions that can be superimposed or combined, reflecting the general solution's structure. The general solution to such an equation involves combining all possible solutions, which are expressible and solvable using the roots of the characteristic equation.
In this context, a linear homogeneous differential equation with constant coefficients yields solutions like:\[z(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}\]where the roots (\(r_1, r_2\)) are related directly to the characteristic equation, \(r_1 = \frac{1}{3}\) and\(r_2 = -\frac{1}{3}\) in our problem.

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Most popular questions from this chapter

When people smoke, carbon monoxide is released into the air. In a room of volume \(60 \mathrm{m}^{3},\) air containing \(5 \%\) carbon monoxide is introduced at a rate of 0.002 \(\mathrm{m}^{3} / \mathrm{min.}\) (This means that \(5 \%\) of the volume of the incoming air is carbon monoxide.) The carbon monoxide mixes immediately with the rest of the air, and the mixture leaves the room at the same rate as it enters. (a) Write a differential equation for \(c(t),\) the concentration of carbon monoxide at time \(t,\) in minutes. (b) Solve the differential equation, assuming there is no carbon monoxide in the room initially. (c) What happens to the value of \(c(t)\) in the long run?

If \(y=e^{2 t}\) is a solution to the differential equation $$ \frac{d^{2} y}{d t^{2}}-5 \frac{d y}{d t}+k y=0 $$ find the value of the constant \(k\) and the general solution to this equation.

\(p\) and \(q\) are the number of individuals in two interacting populations with \(p, q>0\) satisfying the system of equations $$ \begin{array}{l} \frac{1}{p} \frac{d p}{d t}=0.01 q-0.3 \\ \frac{1}{q} \frac{d q}{d t}=0.02 p-0.2 \end{array} $$ What is the relative rate of change of \(p\) if \(q=10 ?\)

Are the statements true or false? Give an explanation for your answer. For any positive values of the constant \(k\) and any positive values of the initial value \(P(0),\) the solution to the differential equation \(d P / d t=k P(L-P)\) has limiting value \(L\) as \(t \rightarrow \infty\).

Is the statement true or false? Assume that \(y=f(x)\) is a solution to the equation \(d y / d x=g(x) .\) If the statement is true, explain how you know. If the statement is false, give a counterexample. If \(g(x)\) is periodic, then \(f(x)\) is also periodic.

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