Question

In Exercises \(1-12\) find the general solution. $$ y^{\prime \prime}+6 y^{\prime}+10 y=0 $$

Short answer

Question: Find the general solution for the second-order linear homogeneous ordinary differential equation \(y^{\prime \prime}+6 y^{\prime}+10 y=0\). Answer: The general solution for the given differential equation is \(y(t) = e^{-3t}(C_1 \cos{(1t)} + C_2 \sin{(1t)})\).

Step-by-step solution

Write the Characteristic Equation

Given the equation \(y^{\prime \prime}+6 y^{\prime}+10 y=0\), we can write the characteristic equation by replacing \(y^{\prime \prime}\) with \(r^2\), \(y^{\prime}\) with \(r\), and \(y\) with 1. The characteristic equation is then: $$ r^2 + 6r + 10 = 0 $$

Find the Roots of the Characteristic Equation

Now, we need to find the roots of the equation \(r^2 + 6r + 10 = 0\). We can solve this quadratic equation by either factoring, completing the square, or using the quadratic formula. Here, we will use the quadratic formula: $$ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ For our equation, \(a = 1\), \(b = 6\), and \(c = 10\). Plugging in these values, we get: $$ r = \frac{-6 \pm \sqrt{6^2 - 4(1)(10)}}{2(1)} $$ Simplifying further, we have: $$ r = \frac{-6 \pm \sqrt{36 - 40}}{2} $$ $$ r = -3 \pm i $$ Since we have complex roots, the general solution is in the form of \(y(t) = e^{αt}(C_1 \cos{βt} + C_2 \sin{βt})\), where α is the real part of the complex roots and β is the imaginary part.

Write the General Solution

Now, we can write the general solution for the given differential equation using the roots we found: \(α = -3\) and \(β = 1\). The general solution is: $$ y(t) = e^{-3t}(C_1 \cos{(1t)} + C_2 \sin{(1t)}) $$

Key concepts

Characteristic Equation

Understanding the characteristic equation is essential when dealing with linear homogeneous second order differential equations. The characteristic equation is a tool that helps us find the solution to differential equations by turning them into algebraic equations.

The process of forming the characteristic equation involves substituting derivatives in the original differential equation with powers of an unknown variable, commonly denoted as 'r'. For example, a second order derivative, such as \(y''\), is replaced by \(r^2\), a first order derivative, such as \(y'\), by \(r\), and a zeroth order term (the function itself, \(y\)) by 1. This reduces the differential equation to a polynomial equation in terms of 'r', which is the characteristic equation we want.

It is crucial to properly understand how to correctly substitute the derivatives and set up this equation, as it paves the way to finding the general solution of the differential equation. The characteristic equation for the given exercise is \(r^2 + 6r + 10 = 0\), succinctly transforming our differential problem into an algebraic one.

Complex Roots

When analyzing the characteristic equation, we may encounter complex roots, which occur when the discriminant (\(b^2 - 4ac\)) of the quadratic equation is negative. Complex roots always come in conjugate pairs, of the form \(a \text{±} bi\), where 'i' denotes the imaginary unit.

In the realm of differential equations, complex roots signal that the solution will involve both exponential and trigonometric functions. This is due to Euler's formula, which connects the trigonometric and exponential functions with complex exponents.

For the differential equation at hand, the roots are \(-3 \text{±} i\), indicating oscillatory behavior damped by an exponential decay because of the negative real part. One should interpret these complex roots as a mandate to write the general solution as a combination of exponential decay, sine, and cosine functions, revealing the inherent oscillation within the system described by the differential equation.

Quadratic Formula

The quadratic formula \(r = \frac{-b \text{±} \text{sqrt}{b^2 - 4ac}}{2a}\) is the silver bullet for finding the roots of any quadratic equation, which is an equation of the form \(ax^2 + bx + c = 0\). When the characteristic equation derived from a second order differential equation is quadratic, the quadratic formula is your go-to method for finding its roots.

Once you identify the coefficients 'a', 'b', and 'c' from the characteristic equation, plug them into the quadratic formula to find the roots. These roots may be real or complex and determine the form of the general solution to the original differential equation. In this exercise, the quadratic formula helped us unearth the complex roots \(-3 \text{±} i\), guiding us towards the correct form for the general solution.

Second Order Differential Equation

A second order differential equation is an equation that involves the unknown function \(y\) and its derivatives up to the second order. It is expressed as \(a_2y''+a_1y'+a_0y=g(t)\), where \(g(t)\) is a given function of time. If \(g(t)\) is zero, as in the exercise, the equation is called homogeneous.

Homogeneous second order differential equations are solved by finding the general solution that satisfies the equation. This process involves the three crucial steps exemplified in the exercise: forming the characteristic equation, finding its roots (using methods such as the quadratic formula), and writing the general solution based on whether the roots are real or complex.

The general solution for a homogeneous second order differential equation with constant coefficients, as shown in the exercise, is a linear combination of functions that include exponentials and possibly sines and cosines, depending on the nature of the roots.