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Chapter 31: Problem 22

Solve the given differential equations. $$5 y^{\prime \prime}-y^{\prime}=3 y$$

Short Answer

Expert verified
The general solution is \(y(t) = C_1 e^{\left(\frac{{1 + \sqrt{61}}}{10}\right) t} + C_2 e^{\left(\frac{{1 - \sqrt{61}}}{10}\right) t}\)."

Step by step solution

01

Recognize the Type of Differential Equation

The given differential equation is \(5 y'' - y' = 3y\). This is a second-order linear differential equation with constant coefficients.
02

Write the Characteristic Equation

To solve the equation, first create the characteristic equation by replacing \(y''\) with \(r^2\), \(y'\) with \(r\), and \(y\) with 1: \(5r^2 - r - 3 = 0\).
03

Solve the Characteristic Equation

Apply the quadratic formula \(r = \frac{{-b \, \pm \,\sqrt{{b^2 - 4ac}}}}{2a}\) where \(a = 5\), \(b = -1\), \(c = -3\). Calculate the discriminant \(\Delta = (-1)^2 - 4 \cdot 5 \cdot (-3) = 1 + 60 = 61\). Thus, \(r = \frac{{1 \, \pm \, \sqrt{61}}}{10}\).
04

Write the General Solution

The roots are real and distinct, so the general solution takes the form \(y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}\) where \(r_1 = \frac{{1 + \sqrt{61}}}{10}\) and \(r_2 = \frac{{1 - \sqrt{61}}}{10}\).
05

Final General Solution

Thus, the general solution of the differential equation is \(y(t) = C_1 e^{\left(\frac{{1 + \sqrt{61}}}{10}\right) t} + C_2 e^{\left(\frac{{1 - \sqrt{61}}}{10}\right) t}\).

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Key Concepts

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

Second-Order Linear Differential Equations
Second-order linear differential equations are a fundamental part of calculus and engineering. These equations involve the second derivative of a function, which means they are concerned with the rate of change of a rate of change. In simpler terms, they describe how acceleration changes over time, which is crucial in fields such as physics and engineering.
  • A second-order differential equation has the general form: \[ a y'' + b y' + c y = f(t) \]where \( a, b, \) and \( c \) are constants, and \( f(t) \) is a function of the independent variable \( t \).
  • These equations can model a variety of physical systems, such as harmonic oscillators (e.g., mass-spring systems) and electrical circuits.
  • The category 'linear' says that the dependent variable \( y \) and all its derivatives appear to the first power and are not multiplied together.
When solving these types of equations, we're usually looking for a function \( y(t) \) that satisfies this relationship expressed by the equation.
Characteristic Equation
When dealing with linear differential equations with constant coefficients, constructing the characteristic equation is a key step. This equation helps us find the solutions to a differential equation in a more manageable form.
  • The characteristic equation transforms the differential equation into a polynomial. For a second-order linear differential equation like the one described, it takes the form: \[ a r^2 + b r + c = 0 \]
  • The process involves substitution, where each derivative corresponds to a power of \( r \):
    • \( y'' \rightarrow r^2 \)
    • \( y' \rightarrow r \)
    • \( y \rightarrow 1 \)
  • This transformation suggests that solving the differential equation can be akin to finding the roots of this polynomial equation.
Once the characteristic equation is established, the solutions or roots \( r \) can point us toward the function \( y(t) \) that solves the original equation.
Quadratic Formula
The quadratic formula is a powerful algebraic tool for solving second-order polynomial equations of the form \( ax^2 + bx + c = 0 \). The roots of the equation are found using:
\[ r = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]
  • The terms \( a, b, \) and \( c \) represent the coefficients of the quadratic equation.
  • The discriminant \( b^2 - 4ac \) plays a crucial role:
    • If positive, there are two distinct real roots.
    • If zero, there is one real root.
    • If negative, the roots are complex or imaginary numbers.
In our specific exercise, the discriminant is 61, a positive number, indicating two distinct real roots for the characteristic equation. Using this information, we derive two possible values for \( r \), which are essential in constructing the general solution of the differential equation.
General Solution
Once the roots of the characteristic equation are determined, you can find the general solution of the second-order linear differential equation. This solution is a crucial part of understanding the behavior of the system described by the differential equation.
  • If the roots are real and distinct, as in this case, the solution is:\[ y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \]where \( C_1 \) and \( C_2 \) are constants determined by initial or boundary conditions.
  • \( r_1 \) and \( r_2 \) are the roots found using the quadratic formula.
  • The exponential functions express how the effect of each root transpires over time.
In summary, when given initial conditions, the constants \( C_1 \) and \( C_2 \) can be found, completing the specific solution to the differential equation. This solution allows us to predict system behavior under diverse conditions.

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Most popular questions from this chapter

Solve the given problems. Explain why more than one function can be a solution of a given differential equation. Find three different functions that are solutions of \(y^{\prime}=y\).

Solve the given problems. In an electric circuit, if a capacitor discharges through a negligible resistance, the current \(i\) is related to the time \(t\) by the equation \(d^{2} i / d t^{2}=-a^{2} i,\) where \(a\) is a constant. Find the frequency of the current if \(a=1000\)

Solve the given differential equations by Laplace transforms. The function is subject to the given conditions. A spring is stretched 1 \(\mathrm{m}\) by a 20-N weight. The spring is stretched \(0.5 \mathrm{m}\) below the equilibrium position with the weight attached and then released. If it is in a medium that resists the motion with a force equal to \(12 v,\) where \(v\) is the velocity, find the displacement \(y\) of the weight as a function of the time.

Solve the given problems. For a given circuit, \(L=0.100 \mathrm{H}, R=0, C=100 \mu \mathrm{F},\) and \(E=100 \mathrm{V} .\) Find the equation relating the charge and the time if \(q=0\) and \(i=0\) when \(t=0\)

Solve the given problems by solving the appropriate differential equation. The rate of change of air pressure \(p\left(\text { in } \mathrm{Ib} / \mathrm{ft}^{2}\right)\) with respect to height \(h\) (in \(\mathrm{ft}\) ) is approximately proportional to the pressure. If the pressure is 15.0 lb/in. \(^{2}\) when \(h=0\) and \(p=10.0\) lb/in. \(^{2}\) when \(h=9800 \mathrm{ft}\) find the expression relating pressure and height.
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