Question

Verify that the given function \(y\) is a solution of the initial value problem that follows it. $$y(t)=-3 \cos 3 t ; y^{\prime \prime}(t)+9 y(t)=0, y(0)=-3, y^{\prime}(0)=0$$

Short answer

Question: Verify that the function \(y(t) = -3\cos{3t}\) is a solution to the initial value problem \(y''(t) + 9y(t) = 0\), with initial conditions \(y(0) = -3\) and \(y'(0) = 0\). Answer: After computing the first and second derivatives of the given function, we found that the equation \(y''(t) + 9y(t) = 0\) is true for all values of \(t\). We also verified that the initial conditions \(y(0) = -3\) and \(y'(0)=0\) are met. Therefore, the given function \(y(t) = -3\cos{3t}\) is a solution to the initial value problem.

Step-by-step solution

Find \(y'(t)\)

To find the first derivative, we will differentiate \(y(t)\) with respect to \(t\): $$y(t) = -3\cos{3t}$$ $$y'(t) = \frac{d}{dt}(-3\cos{3t}) = -3\left(-3\sin{3t}\right) = 9\sin{3t}.$$ Step 2: Find the second derivative of \(y(t)\).

Find \(y''(t)\)

To find the second derivative, we will differentiate \(y'(t)\) with respect to \(t\): $$y'(t) = 9\sin{3t}$$ $$y''(t) = \frac{d}{dt}(9\sin{3t}) = 9(3\cos{3t}) = 27\cos{3t}.$$ Step 3: Substitute the function and its derivatives into the given equation.

Substitute \(y(t)\), \(y'(t)\), and \(y''(t)\)

Now we will check if the equation \(y''(t) + 9y(t) = 0\) holds true: $$27\cos{3t} + 9(-3\cos{3t}) = 27\cos{3t} - 27\cos{3t} = 0.$$ The equation is true for all values of \(t\). Step 4: Check the initial conditions.

Checking initial conditions

We need to verify if the given initial conditions \(y(0) = -3\) and \(y'(0) = 0\) hold true: For \(y(0) = -3\), $$y(0) = -3\cos{(3\times 0)} = -3\cos{0} = -3.$$ For \(y'(0) = 0\), $$y'(0) = 9\sin{(3\times 0)} = 9\sin{0} = 0.$$ Both initial conditions are met. Therefore, we can conclude that \(y(t) = -3\cos{3t}\) is a solution to the given initial value problem.

Key concepts

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. These equations are fundamental in capturing how things change, which can apply to various fields like physics, engineering, and economics. In this particular problem, we are given the differential equation:
\[y''(t) + 9y(t) = 0\]This is a second-order linear homogeneous differential equation. "Second-order" because it involves a second derivative \(y''(t)\), and "homogeneous" because it equals zero. Such equations often arise in systems experiencing simple harmonic motion, responsible for describing oscillations like those of pendulums or springs. The task is to find functions like \(y(t) = -3\cos{3t}\) that, when substituted back into the equation along with their derivatives, fulfill the equation perfectly.
Understanding these types of equations helps students explore the core concepts of changes and trends in the world around them. It shows how mathematical principles can predict and elucidate natural phenomena.

Solution Verification

Solution verification is an essential step when working with differential equations. It involves checking if the proposed solution satisfies the equation and any initial conditions.
To verify a solution, we substitute the function and its derivatives back into the original differential equation to see if the equation holds true. Here, we calculated the first derivative \(y'(t) = 9\sin{3t}\) and the second derivative \(y''(t) = 27\cos{3t}\) of the function \(y(t) = -3\cos{3t}\).
After substituting, the calculation showed:

• \(y''(t) + 9y(t) = 27\cos{3t} - 27\cos{3t} = 0\)

This verification step confirmed that the equation balances to zero, thus proving that the differential equation is satisfied by the proposed function.
Moreover, verifying a solution develops a deeper intuitive understanding of both the solution and the behavior it models.

Initial Conditions

Initial conditions are specific values that the solution of a differential equation must satisfy. These conditions are stated at the outset and give additional information needed to determine the unique solution to a problem.
For the current exercise, the initial conditions are:

• \(y(0) = -3\)
• \(y'(0) = 0\)

These were checked by evaluating the function and its derivative at \(t = 0\). We substitute \(t = 0\) into \(y(t) = -3\cos{3t}\) and find that it equals \(-3\), satisfying the first initial condition. Next, we check \(y'(t)\) at \(t = 0\), which results in \(0\), thus satisfying the second.
Initial conditions are crucial because, while a differential equation can have many solutions, these conditions ensure that only one solution fits the physical or real-world context of the problem.