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Find the general solution to the given differential equation. $$4 z^{\prime \prime}+8 z^{\prime}+3 z=0$$

Short Answer

Expert verified
The general solution is \(z(t) = C_1 e^{-\frac{1}{2} t} + C_2 e^{-\frac{3}{2} t}\).

Step by step solution

01

Identify the type of differential equation

The given differential equation is \(4 z'' + 8 z' + 3 z = 0\), which is a second-order linear homogeneous differential equation with constant coefficients.
02

Write the characteristic equation

Convert the differential equation into its characteristic form. For \(a z'' + b z' + cz = 0\), the characteristic equation is \(a r^2 + b r + c = 0\). Hence, for our equation, it is \(4 r^2 + 8 r + 3 = 0\).
03

Solve the characteristic equation

Solve the quadratic equation \(4 r^2 + 8 r + 3 = 0\) using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). For this equation, \(a = 4\), \(b = 8\), \(c = 3\).
04

Calculate the discriminant

First, calculate the discriminant: \(b^2 - 4ac = 8^2 - 4(4)(3) = 64 - 48 = 16\). The discriminant is positive, indicating two distinct real roots.
05

Find the roots using the quadratic formula

Substituting values into the quadratic formula gives \(r = \frac{-8 \pm \sqrt{16}}{8}\). Simplifying, the roots are \(r_1 = -\frac{1}{2}\) and \(r_2 = -\frac{3}{2}\).
06

Write the general solution

For distinct real roots \(r_1\) and \(r_2\), the general solution of the differential equation is \(z(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}\). Substitute the roots: \(z(t) = C_1 e^{-\frac{1}{2} t} + C_2 e^{-\frac{3}{2} t}\).

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Key Concepts

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

Second-Order Differential Equations
A second-order differential equation involves a second derivative, which for this exercise is denoted as \( z'' \). These types of equations are often used to describe physical systems, like motion under the influence of force, electrical circuits, or mechanical vibrations. A second-order differential equation typically has the form \( a y'' + b y' + c y = f(t) \), where \( y'' \) is the second derivative of the unknown function \( y \), \( y' \) is the first derivative, and \( f(t) \) is a given function, which is zero in our homogeneous case. The given differential equation, \( 4z'' + 8z' + 3z = 0 \), is a classic example of a second-order differential equation that can be solved using characteristic methods when the coefficients are constant.
Characteristic Equation
The characteristic equation is derived from the given differential equation by replacing the derivatives with powers of \( r \). This transformation simplifies finding solutions by turning the problem into a polynomial equation. For a standard form equation \( a y'' + b y' + c y = 0 \), the characteristic equation is \( a r^2 + b r + c = 0 \).

For our particular equation, \( 4z'' + 8z' + 3z = 0 \), it translates into \( 4r^2 + 8r + 3 = 0 \). Solving this equation uses the quadratic formula, \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which solves for \( r \) based on the coefficients \( a \), \( b \), and \( c \). The roots determined through the characteristic equation are essential in constructing the solution to the differential equation.
Homogeneous Differential Equations
Homogeneous differential equations have a specific definition where the equation is set to equal zero, meaning there are no external forces or functions acting upon the system described by the equation. This property simplifies solving techniques. For the equation \( 4z'' + 8z' + 3z = 0 \), the term on the right-hand side is zero, confirming its status as homogeneous.

Homogeneity in this context implies that the solution relies solely on the intrinsic properties of the equation rather than any external input functions. The solutions we find will form a family of curves, each differing by constants, which arise from the integration process during solving.
General Solution
The goal of solving a differential equation like \( 4z'' + 8z' + 3z = 0 \) is to find the general solution. This solution encompasses all possible specific solutions, represented by arbitrary constants that adjust based on initial conditions.

In the case of second-order linear homogeneous differential equations with distinct real roots, the general solution is given by \( z(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \). Here, \( C_1 \) and \( C_2 \) are constants determined by initial or boundary conditions, and \( r_1 \) and \( r_2 \) are the roots found from the characteristic equation.

Using the roots \( r_1 = -\frac{1}{2} \) and \( r_2 = -\frac{3}{2} \), the specific general solution for our exercise is \( z(t) = C_1 e^{-\frac{1}{2}t} + C_2 e^{-\frac{3}{2}t} \), which describes how the function \( z \) evolves over time, influenced only by starting conditions.

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Most popular questions from this chapter

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