Question

Finding a Particular Solution In Exercises \(35-40\) , verify that the general solution satisfies the differential equation. Then find the particular solution that satisfies the initial condition(s). $$ \begin{array}{l}{y=C_{1} \sin 3 x+C_{2} \cos 3 x} \\ {y^{\prime \prime}+9 y=0} \\ {y=2 \text { when } x=\frac{\pi}{6}} \\ {y^{\prime}=1 \text { when } x=\frac{\pi}{6}}\end{array} $$

Short answer

The differential equation \(y^{\prime \prime}+9y=0\)'s particular solution that satisfies the initial condition, is found to be \(y=2 \sin 3x + \frac{1}{3} \cos 3x\).

Step-by-step solution

Verify the General Solution

First, we check if the general solution \(y=C_{1} \sin 3x + C_{2} \cos 3x\) satisfies the differential equation \(y^{\prime \prime}+9y=0\). To do so, we calculate the second derivative of y, which is \(-9C_{1} \sin 3x - 9C_{2} \cos 3x\). Adding 9 times \(y\) to this yields zero, thus confirming that the general solution satisfies the provided differential equation.

Find the Constant \(C_1\) and \(C_2\)

To find the constants \(C_1\) and \(C_2\), we use the initial conditions. Substituting \(x=\frac{\pi}{6}\) in the original equation gives us \(y=C_{1} \sin \frac{\pi}{2} + C_{2} \cos \frac{\pi}{2} = C_{1}\). From the condition \(y=2\) when \(x=\frac{\pi}{6}\), we find that \(C_{1}=2\). Using the same \(x=\frac{\pi}{6}\) in the derivative of the original solution \(y^{'}=3C_{1} \cos 3x - 3C_{2} \sin 3x\), and setting it equal to 1 (provided in the problem), we get \(3C_{2} = 1\), which gives \(C_{2} = \frac{1}{3}\).

Write out the Solution

Substitute the values of \(C_1\) and \(C_2\) we obtained in the second step into our general solution. The particular solution will thus be \(y=2 \sin 3x + \frac{1}{3} \cos 3x\).

Key concepts

Particular Solution

In the context of differential equations, a "particular solution" refers to a solution that satisfies both the differential equation and the given initial conditions. A particular solution is specific to the problem at hand, tailored by using the initial conditions provided in the exercise.

In this particular exercise, we started with a general solution, which is written as \( y = C_1 \sin 3x + C_2 \cos 3x \). This solution solves the differential equation \( y'' + 9y = 0 \) generally, meaning it could represent any solution curve that satisfies the equation. However, to pinpoint one specific curve out of potentially many, we need to satisfy the initial conditions given in the problem.

• "Initial conditions" refer to known values such as \( y = 2 \) and \( y' = 1 \) when \( x = \frac{\pi}{6} \).
• These conditions allow us to determine specific values for constants \( C_1 \) and \( C_2 \), ultimately leading to the specific, or particular, solution: \( y = 2 \sin 3x + \frac{1}{3} \cos 3x \).

General Solution

A "general solution" is a solution to a differential equation that contains arbitrary constants, representing a family of solutions. In our problem, the general solution \( y = C_1 \sin 3x + C_2 \cos 3x \) includes two constants, \( C_1 \) and \( C_2 \). These constants account for the innumerable solution curves that the differential equation could represent.

The process begins by ensuring the general solution satisfies the equation. This involves:

• Calculating derivatives as required by the differential equation, such as \( y'' \).
• Substituting these into the equation to verify equality.

For our equation \( y'' + 9y = 0 \), substituting \( y'' = -9C_1 \sin 3x - 9C_2 \cos 3x \) results in zero when combined with \( 9y \). This verifies that our general solution is valid across all values of \( C_1 \) and \( C_2 \). Remember, the transition from a general solution to particular involves using provided conditions, which we calculate next.

Initial Conditions

Initial conditions are specific, known values provided in differential equations to determine particular solutions from a general solution. They typically include values for the function and possibly its derivatives at certain points, such as the ones given in our exercise: \( y = 2 \) and \( y' = 1 \) when \( x = \frac{\pi}{6} \).

These conditions are critical because:

• They allow us to solve for arbitrary constants in the general solution.
• They ensure the particular solution fits specific criteria, essentially situating the solution in "real-world" context.

In our exercise, by applying the initial conditions, we substitute \( y \) and \( y' \) into the general solution and its derivative. This allows us to find \( C_1 = 2 \) and \( C_2 = \frac{1}{3} \), which are crucial in forming the specific solution curve that not only satisfies the differential equation but also fits the scenario outlined by the problem's constraints.

Second Derivative

The second derivative represents the rate of change of the rate of change of a function and plays a vital role in solving differential equations, especially those involving second-order equations like our problem. In essence, it helps in describing how the slope of a function is changing at any given point.

To solve the differential equation \( y'' + 9y = 0 \), we need the second derivative \( y'' \) of the general solution \( y = C_1 \sin 3x + C_2 \cos 3x \).

• First, calculate the first derivative: \( y' = 3C_1 \cos 3x - 3C_2 \sin 3x \).
• Then, derive it again to get the second derivative: \( y'' = -9C_1 \sin 3x - 9C_2 \cos 3x \).
• Substitute \( y'' \) and \( y \) back into the differential equation to ensure it satisfies the equality, confirming the general solution is correct.

Understanding and calculating the second derivative is essential because it allows us to confirm the general solution mathematically fits the criteria set by the differential equation. It's one key step in narrowing down to the particular solution by applying the specific initial conditions.