Question

In Exercises \(13-17\) solve the initial value problem. $$ 6 y^{\prime \prime}-y^{\prime}-y=0, \quad y(0)=10, \quad y^{\prime}(0)=0 $$

Short answer

Answer: The specific solution for \(y(t)\) is \(y(t) = 6 e^{\frac{1}{2}t} + 4 e^{-\frac{1}{3}t}\).

Step-by-step solution

Find the characteristic equation

From the given differential equation \(6y^{\prime\prime} - y^{\prime} - y=0\), let's find the characteristic equation by considering \(y(t)=e^{rt}\), where r is the root. Using this, we can rewrite the given equation: $$ 6r^2 e^{rt} - re^{rt} - e^{rt}=0 $$ Since \(e^{rt} \neq 0\), we can divide the equation by \(e^{rt}\), resulting in the characteristic equation: $$ 6r^2 - r - 1 = 0 $$

Solve the characteristic equation

Now, we need to solve the quadratic equation \(6r^2 - r - 1 = 0\). By applying the quadratic formula, we get: $$ r = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(6)(-1)}}{2(6)} = \frac{1 \pm \sqrt{1 + 24}}{12} $$ Thus, we have: $$ r_{1,2} = \frac{1 \pm \sqrt{25}}{12} = \frac{1 \pm 5}{12} $$ Giving us the roots, \(r_1 = \frac{1+5}{12} = \frac{1}{2}\) and \(r_2 = \frac{1-5}{12} = -\frac{1}{3}\).

Write the general solution

Based on the roots obtained, the general solution for this differential equation can be written as: $$ y(t) = C_1 e^{\frac{1}{2}t} + C_2 e^{-\frac{1}{3}t} $$ Where \(C_1\) and \(C_2\) are constants.

Determine the specific solution

Now, we need to find the specific solution using the initial conditions given: \(y(0) = 10\) and \(y^{\prime}(0) = 0\). We will use these conditions to find the values of \(C_1\) and \(C_2\). First, let's apply \(y(0) = 10\): $$ 10 = C_1 e^{0} + C_2 e^{0} \Rightarrow 10 = C_1 + C_2 $$ Next, let's apply \(y^{\prime}(0) = 0\). For this, we need to find the first derivative of y(t): $$ y^{\prime}(t) = \frac{1}{2}C_1 e^{\frac{1}{2}t} - \frac{1}{3}C_2 e^{-\frac{1}{3}t} $$ Now, apply the initial condition \(y^{\prime}(0) = 0\): $$ 0 = \frac{1}{2}C_1 e^{0} - \frac{1}{3}C_2 e^{0} \Rightarrow 0 = \frac{1}{2}C_1 - \frac{1}{3}C_2 $$ Solving the system of equations, we find the constants: \(C_1 = 6\) and \(C_2 = 4\).

Final specific solution

Now that we have found \(C_1\) and \(C_2\), we can write the specific solution for the given IVP: $$ y(t) = 6 e^{\frac{1}{2}t} + 4 e^{-\frac{1}{3}t} $$ This is the final solution for the given initial value problem.

Key concepts

Characteristic Equation

When solving linear differential equations, one of the critical steps involves finding the characteristic equation. This equation emerges from substituting a trial solution of the form y(t) = e^{rt} into a homogeneous linear differential equation, assuming that e^{rt} ≠ 0 for all values of t. This substitution is helpful because the exponential function's derivatives are proportional to the function itself, simplifying the algebra.

The resulting characteristic equation is typically a polynomial whose roots will give us information about the behavior of the differential equation's solutions. In the context of our problem, after simplifying the differential equation with the trial solution, we obtained a quadratic characteristic equation 6r^2 - r - 1 = 0, which will be solved using the quadratic formula to find the roots. These roots are pivotal as they are directly used in formulating the general solution of the differential equation.

Differential Equations

A differential equation is a mathematical equation that relates some function with its derivatives. In applications, the functions generally represent physical quantities, the derivatives represent their rates of change, and the differential equation defines a relationship between the two. These equations are fundamental in describing various phenomena such as motion, heat, electricity, and fluid flow.

The exercise we're discussing involves a second-order linear homogeneous differential equation with constant coefficients, which is a common type found in physics and engineering. The given initial conditions allow us to find a specific solution, which makes the problem an initial value problem (IVP). Explaining how solutions to such problems are located by obtaining the characteristic equation and solving it allows learners to understand the systematic approach to solving differential equations.

Quadratic Formula

An indispensable tool in algebra is the quadratic formula. It is used for solving quadratic equations, which are polynomials of the second degree. The general form of a quadratic equation is ax^2 + bx + c = 0, and the quadratic formula provides the solutions as x = (-b ± √(b^2 - 4ac))/(2a).

The discriminant, b^2 - 4ac, within the quadratic formula, decides the nature of the roots (real and distinct, real and repeated, or complex). In the context of solving the characteristic equation for the differential equation, using the quadratic formula helped us find two distinct real roots, which leads directly to constructing the general solution. This solidifies the connection between algebraic methods, such as the quadratic formula, and their use in solving calculus problems like differential equations.

General Solution

The 'general solution' to a differential equation represents a family of functions that includes every possible solution to the differential equation. In the case of linear differential equations with constant coefficients, the general solution typically comprises a combination of exponentials, sines, and cosines, with arbitrary constants.

The roots of the characteristic equation help in determining the form of this general solution. For instance, real distinct roots lead to exponential terms, as seen in our exercise, where the roots resulted in a general solution of the form y(t) = C_1e^{rt1} + C_2e^{rt2} with C_1 and C_2 as arbitrary constants. To transform this general solution into a specific one that satisfies the given initial conditions, we use the initial values to determine the exact values of these constants. This illustrates how general solutions are adapted to meet the specific constraints of an initial value problem.