Question

In Exercises \(18-21\) solve the initial value problem and graph the solution. $$ \text { C/G } y^{\prime \prime}+4 y^{\prime}+10 y=0, \quad y(0)=3, \quad y^{\prime}(0)=-2 $$

Short answer

Answer: The particular solution to the given initial value problem is: $$ y(t) = e^{-2t}(3\cos(2\sqrt{6}t)+\frac{2}{\sqrt{6}}\sin(2\sqrt{6}t)) $$ The graph of this solution will represent a damped oscillation where the oscillations decay as time progresses.

Step-by-step solution

Characteristic Equation

First, let's rewrite the given equation in operator form: $$ (C/G)D^2y + 4Dy + 10y = 0 $$ Now, let's form the characteristic equation by replacing the operators \((D^2)\) and \((D)\) with \(m\). $$ (C/G)m^2 + 4m + 10 = 0 $$

Find Roots of the Characteristic Equation

The characteristic equation is a quadratic equation. Let's find its roots using the quadratic formula: $$ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where \(a=\frac{C}{G}\), \(b=4\), and \(c=10\). Plugging in the values, we get: $$ m = \frac{-4 \pm \sqrt{4^2 - 4\left(\frac{C}{G}\right)(10)}}{2\left(\frac{C}{G}\right)} $$ Since we are dealing with a homogeneous equation, \(C/G\) is a constant multiple that won't affect the roots. Therefore, we can just assume that \(C/G=1\). So the equation becomes: $$ m = \frac{-4 \pm \sqrt{4^2 - 4(10)}}{2} $$ Calculating the discriminant, we notice it is negative. $$ \Delta = 4^2 - 4(10) = 16 - 40 = -24 < 0 $$ This means we have two complex roots. Let's compute these, $$ m_1, m_2 = -2 \pm \frac{\sqrt{-24}}{2} = -2 \pm \frac{\sqrt{24}i}{2} = -2 \pm 2i\sqrt{6} $$

Write General Solution

Since the roots are complex, the general solution can be written in the form: $$ y(t) = e^{-2t}(c_1\cos(2\sqrt{6}t)+c_2\sin(2\sqrt{6}t)) $$

Find Particular Solution Using Initial Conditions

Now, let's use the initial conditions to find the particular solution. 1. \(y(0) = 3\): Plugging \(t=0\) into the general solution, we get $$ 3 = e^0(c_1\cos(0)+c_2\sin(0)) = c_1 $$ 2. \(y'(0) = -2\): First, let's find the derivative of the general solution: $$ y'(t) = -2e^{-2t}(c_1\cos(2\sqrt{6}t)+c_2\sin(2\sqrt{6}t)) - e^{-2t}(c_1(-2\sqrt{6}\sin(2\sqrt{6}t))+c_2(2\sqrt{6}\cos(2\sqrt{6}t))) $$ Now, plugging \(t=0\), we get $$ -2 = -2(c_1\cos(0)+c_2\sin(0)) - (c_1(-2\sqrt{6}\sin(0))+c_2(2\sqrt{6}\cos(0))) $$ Since \(c_1 = 3\) and \(\sin(0)=0, \cos(0)=1\), this simplifies to $$ -2 = -2(3)+ 2\sqrt{6}c_2 \Rightarrow c_2 = \frac{-2 + 6}{2\sqrt{6}} = \frac{2}{\sqrt{6}} $$ Therefore, the particular solution is: $$ y(t) = e^{-2t}(3\cos(2\sqrt{6}t)+\frac{2}{\sqrt{6}}\sin(2\sqrt{6}t)) $$

Sketch the Graph of the Solution

To graph the solution, you can use a graphing calculator or software (such as Desmos, GeoGebra, or WolframAlpha) to plot the function: $$ y(t) = e^{-2t}(3\cos(2\sqrt{6}t)+\frac{2}{\sqrt{6}}\sin(2\sqrt{6}t)) $$ The graph will display a damped oscillating function (since roots are complex and the damping coefficient is negative). The amplitude will decrease as the time increases, which means that the oscillations will decay as the time progresses.

Key concepts

Characteristic Equation

In the context of solving differential equations, a characteristic equation plays a vital role. It is derived from the original differential equation by replacing the differential operators with a designated variable, often denoted as \( m \). This transformation allows us to simplify the problem into an algebraic equation.
- The characteristic equation typically takes the form of a polynomial, often quadratic when dealing with second-order linear differential equations.
- Solving this polynomial provides us with the roots, which tell us the type of solutions the differential equation has.
For example, in the exercise, the given equation was transformed into the characteristic equation \( m^2 + 4m + 10 = 0 \). This equation guides us towards finding the nature of the solutions, whether they are real or complex.

Complex Roots

Complex roots arise when the discriminant of our characteristic equation is negative. The discriminant, found in quadratic equations, is given by \( b^2 - 4ac \). When it is less than zero, it indicates that the roots are complex.
- Complex roots are usually expressed in the form \( m = \alpha \pm i\beta \). Here, \( \alpha \) and \( \beta \) are real numbers, and \( i \) represents the imaginary unit.
- These roots signal that our solutions involve oscillatory behavior, often combining exponential decay or growth with oscillatory sine and cosine functions.
In our example, we found the roots to be \( -2 \pm 2i\sqrt{6} \), indicating an exponential decay combined with oscillations. This results in a solution featuring these combined behaviors.

General Solution

The general solution of a differential equation is a complete representation of all possible solutions. It typically involves constants that can be determined using initial conditions.
- For equations with complex roots, the general solution is represented using exponential functions multiplied by sine and cosine terms. - This accounts for both the decay (or growth) and oscillatory nature of the system described by the differential equation.
In this exercise, the general solution was expressed as \( y(t) = e^{-2t}(c_1\cos(2\sqrt{6}t) + c_2\sin(2\sqrt{6}t)) \), showcasing the decay (\( e^{-2t} \)) and the oscillation terms.

Initial Conditions

Initial conditions are the values provided at the onset of the problem. They are essential to determining the specific solution out of the general family of solutions by assigning values to the arbitrary constants in the general solution.
- Typically, initial conditions are given as values of the function and its derivatives at a specific point, often \( t = 0 \).
- By plugging these initial conditions into the general solution, we can solve for the constants.
In this problem, initial conditions \( y(0) = 3 \) and \( y'(0) = -2 \) were used to find the constants \( c_1 \) and \( c_2 \). Specifically, solving these led to \( c_1 = 3 \) and \( c_2 = \frac{2}{\sqrt{6}} \), thus deriving the particular solution.