Problem 14 - Advanced Engineering Mathematics

Question

In Problems 13-28, use the procedures developed in this chapter to find the general solution of each differential equation. $$ 2 y^{\prime \prime}+2 y^{\prime}+3 y=0 $$

Step-by-Step Solution

Answer

The general solution is \(y(t) = e^{-\frac{1}{2} t}(C_1 \cos\left(\frac{\sqrt{5}}{2} t\right) + C_2 \sin\left(\frac{\sqrt{5}}{2} t\right))\).

Step 1: Identify the Type of Differential Equation

The given equation is a second-order linear homogeneous differential equation: \(2y'' + 2y' + 3y = 0\). Recognizing the type of equation helps us decide the procedure to find the general solution.

Step 2: Write the Characteristic Equation

For a linear homogeneous differential equation such as \(ay'' + by' + cy = 0\), the characteristic equation is obtained by assuming solutions of the form \(y = e^{rt}\). Substituting into the differential equation gives \(ar^2 + br + c = 0\). For our equation, \(2r^2 + 2r + 3 = 0\) is the characteristic equation.

Step 3: Solve the Characteristic Equation

To solve \(2r^2 + 2r + 3 = 0\), use the quadratic formula: \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 2\), \(b = 2\), and \(c = 3\). The discriminant \(b^2 - 4ac = 2^2 - 4 \cdot 2 \cdot 3 = 4 - 24 = -20\), which is negative, indicating complex roots.

Step 4: Calculate the Complex Roots

Since the discriminant is negative, the roots are complex numbers. Using the formula, \(r = \frac{-2 \pm \sqrt{-20}}{4} = \frac{-2 \pm 2i\sqrt{5}}{4} = \frac{-1 \pm i\sqrt{5}}{2}\). Thus, the roots are \(r_1 = \frac{-1 + i\sqrt{5}}{2}\) and \(r_2 = \frac{-1 - i\sqrt{5}}{2}\).

Step 5: Formulate the General Solution

For complex roots \(r_1 = \alpha + \beta i\) and \(r_2 = \alpha - \beta i\), the general solution of the differential equation is \(y(t) = e^{\alpha t}(C_1 \cos(\beta t) + C_2 \sin(\beta t))\). Using \(\alpha = -\frac{1}{2}\) and \(\beta = \frac{\sqrt{5}}{2}\), the solution becomes \(y(t) = e^{-\frac{1}{2} t}(C_1 \cos\left(\frac{\sqrt{5}}{2} t\right) + C_2 \sin\left(\frac{\sqrt{5}}{2} t\right))\).

Key Concepts

Second-Order Linear Homogeneous Differential Equations

A second-order linear homogeneous differential equation is an important class of differential equations. These equations are in the form of: \[ a y'' + b y' + c y = 0 \] This type of equation involves derivatives of order two, with constant coefficients \(a\), \(b\), and \(c\). "Homogeneous" means there is no term independent of \(y\) on the right-hand side of the equation, notably making the equation equal to zero. Understanding this concept is crucial as it dictates the methods we use to find solutions. How to identify a second-order linear homogeneous differential equation:

• Check for the highest derivative, which should be second order (\(y''\)).
• Ensure all terms are multiplied by something involving \(y\), \(y'\), or \(y''\).
• Verify there are no external or forced terms (like constants or functions of \(x\)).

Recognizing the type of equation aids us in constructing a strategy to solve it effectively.

Characteristic Equation

The characteristic equation is a critical tool for solving homogeneous linear differential equations. It emerges when we assume solutions of the differential equation can take an exponential form: \[ y = e^{rt} \] Substitute this form into the given differential equation, replacing each derivative. You get an algebraic equation related to \(r\): \[ ar^2 + br + c = 0 \] This equation is called the characteristic equation. It transforms the differential equation problem into an algebra problem. This simplification is because exponential functions are solutions to linear differential equations, giving us a straightforward method to explore potential solutions. Practical steps to derive the characteristic equation:

• Replace each derivative using \(y = e^{rt}\).
• Simplify to obtain a quadratic equation in terms of \(r\).
• Solve this quadratic equation to find possible values for \(r\).

This process is foundational as it links differential equations to algebraic processes, making complex problems manageable.

Complex Roots

When solving the characteristic equation, you might encounter complex roots. These arise when the discriminant of the quadratic equation \(b^2 - 4ac\) is negative. This situation implies the roots are in the form: \[ r = \frac{-b \, \pm \, i \sqrt{|b^2 - 4ac|}}{2a} \] Complex roots indicate oscillatory behavior in the system described by the differential equation. Here, the solution involves both cosine and sine functions, reflecting these oscillations. How to interpret and use complex roots:

• Recognize that the real part \(\alpha\) of the root causes exponential decay or growth depending on its sign.
• The imaginary part \(\beta\) results in oscillatory behavior captured by trigonometric functions (\(\cos\) and \(\sin\)).
• Form solutions using complex roots: \( y(t) = e^{\alpha t}(C_1 \cos(\beta t) + C_2 \sin(\beta t)) \).

Understanding complex roots is essential for analyzing systems prone to vibrational modes or damping.

General Solution

The general solution of a second-order linear homogeneous differential equation provides the complete set of solutions. When roots of the characteristic equation are complex, the general solution takes on a unique form embodying both exponential and oscillatory behavior: \[ y(t) = e^{\alpha t}(C_1 \cos(\beta t) + C_2 \sin(\beta t)) \] This expression is built using the real and imaginary parts of the roots. Here's how the solution fits together:

• \(e^{\alpha t}\) represents the exponential part, which dictates growth or decay.
• \(C_1\) and \(C_2\) are arbitrary constants, determined by initial conditions.
• The trigonometric terms handle the oscillatory components, due to the presence of \(\beta\).

This combination of exponential and trigonometric functions allows the general solution to accommodate a wide range of initial conditions and behaviors that depend on the nature of the system being analyzed. Understanding this form equips you to interpret and predict the system's response to various scenarios.