Question

In Exercises 23-28 use a method suggested by Exercise 22 to solve the initial value problem. $$ 9 y^{\prime \prime}+6 y^{\prime}+y=0, \quad y(2)=2, \quad y^{\prime}(2)=-\frac{14}{3} $$

Short answer

The solution to the initial value problem is: $$ y(x) = -4e^{\frac{2}{3}}e^{-\frac{x}{3}} + 6e^{-x} $$

Step-by-step solution

Find the characteristic equation

Write down the characteristic equation for the given ODE. Replace \(y''\) with \(r^2\), \(y'\) with \(r\), and \(y\) with \(1\). The characteristic equation will be: $$ 9r^2 + 6r + 1 = 0 $$

Solve the characteristic equation

Solve the quadratic equation \(9r^2 + 6r + 1 = 0\) to find the roots. Use the quadratic formula: $$ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Plugging in the values, we get: $$ r = \frac{-6 \pm \sqrt{6^2 - 4(9)(1)}}{2(9)} $$ After simplifying, we find the roots are \(r_1 = -\frac{1}{3}\) and \(r_2 = -1\).

Write down the general solution

Since the roots are real and distinct, the general solution is of the form: $$ y(x) = c_1 e^{-\frac{x}{3}} + c_2 e^{-x} $$

Apply the initial conditions

We need to find the values of \(c_1\) and \(c_2\) so that the initial conditions \(y(2) = 2\) and \(y'(2) = -\frac{14}{3}\) are satisfied. First, apply the initial condition \(y(2) = 2\): $$ 2 = c_1 e^{-\frac{2}{3}} + c_2 e^{-2} $$ Next, find the derivative of \(y(x)\): $$ y'(x) = -\frac{1}{3} c_1 e^{-\frac{x}{3}} - c_2 e^{-x} $$ Now apply the other initial condition \(y'(2) = -\frac{14}{3}\): $$ -\frac{14}{3} = -\frac{1}{3} c_1 e^{-\frac{2}{3}} - c_2 e^{-2} $$

Solve for \(c_1\) and \(c_2\)

We have a system of two linear equations with two unknowns, \(c_1\) and \(c_2\): $$ \begin{cases} 2 = c_1 e^{-\frac{2}{3}} + c_2 e^{-2} \\ -\frac{14}{3} = -\frac{1}{3} c_1 e^{-\frac{2}{3}} - c_2 e^{-2} \end{cases} $$ Solve this system of equations for \(c_1\) and \(c_2\). One way to do this is to multiply the second equation by 3 and then subtract the first equation from it: $$ c_1 e^{-\frac{2}{3}} = -4 $$ Solve for \(c_1\): $$ c_1 = -4e^{\frac{2}{3}} $$ Plug the value of \(c_1\) back into the first equation to find \(c_2\): $$ 2 = (-4e^{\frac{2}{3}})e^{-\frac{2}{3}} + c_2 e^{-2} $$ Solve for \(c_2\): $$ c_2 = 6 $$

Write down the final solution

Plug the values of \(c_1\) and \(c_2\) back into the general solution to get the final solution: $$ y(x) = -4e^{\frac{2}{3}}e^{-\frac{x}{3}} + 6e^{-x} $$ This is the solution to the initial value problem.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve derivatives of a function. They play a significant role in modeling real-world phenomena where change is involved, such as motion, growth, decay, and much more. A differential equation relates a function with one or more of its derivatives. In the equation given in the exercise, \(9y'' + 6y' + y = 0\), we see a second-order differential equation. Here:

• \(y''\) is the second derivative of the function \(y\)
• \(y'\) is the first derivative
• \(y\) is the original function with respect to the variable

The task is to find the function \(y(x)\) that satisfies this equation and the given initial conditions. Differential equations like this one often require specific methods for solving, especially when paired with initial conditions.

Characteristic Equation

To find the solution to a linear homogeneous differential equation like the one in the problem, we use the characteristic equation. This technique involves assuming a solution of the form \(y = e^{rx}\) and deriving a quadratic equation whose roots reveal essential features about the solution. For our specific problem, replacing \(y''\), \(y'\), and \(y\) gives us:
\[9r^2 + 6r + 1 = 0\] This quadratic equation, known as the **characteristic equation**, helps in determining the form of the solution:

• If the roots are real and distinct, as they are in this case, the solution comprises exponential terms involving these roots.
• If the roots were equal or complex, the solution would take a different form, involving multiplicative or oscillatory components, respectively.

Solving this characteristic equation reveals the roots \(r_1 = -\frac{1}{3}\) and \(r_2 = -1\), which directly inform the general solution of our differential equation.

Initial Conditions

Initial conditions are the values specified for the function and its derivatives at a particular point. They are crucial as they enable us to find the particular solution that satisfies the specific scenario, making the solution unique.
In the given problem, the initial conditions are:

• \(y(2) = 2\)
• \(y'(2) = -\frac{14}{3}\)

These conditions mean that when \(x = 2\), the function \(y(x)\) is 2, and its derivative is -4.6667 approximately. By substituting these conditions into the general solution derived from the characteristic equation, we solve for constants \(c_1\) and \(c_2\), ensuring that the solution is specifically tailored to these initial values. This step is critical in constructing the unique solution to the initial value problem.

General Solution

The general solution of a differential equation encompasses all possible solutions before applying any initial conditions. When we have a second-order linear differential equation like \(9y'' + 6y' + y = 0\), the general solution is determined based on the roots of its characteristic equation.
In this problem, because the roots were real and distinct, the general solution takes the form:

• \(y(x) = c_1 e^{r_1 x} + c_2 e^{r_2 x}\)

For our specific case, substituting the roots \(r_1 = -\frac{1}{3}\) and \(r_2 = -1\) gives:
\[y(x) = c_1 e^{-\frac{x}{3}} + c_2 e^{-x}\]This expression represents the multitude of potential solutions that our original differential equation could have. However, without the incorporation of the initial conditions, it remains an incomplete solution. By applying these initial conditions, we pin down the values of \(c_1\) and \(c_2\), thus arriving at the specific or particular solution to the given initial value problem.