Question

The function \(y(t)\) is a solution of the initial value problem \(y^{\prime \prime}+a y^{\prime}+b y=0, y\left(t_{0}\right)=y_{0}\), \(y^{\prime}\left(t_{0}\right)=y_{0}^{\prime}\), where the point \(t_{0}\) is specified. Determine the constants \(a, b, y_{0}\), and \(y_{0}^{\prime} .\) $$ y(t)=\sqrt{3} \cos \pi t-\sin \pi t, \quad t_{0}=1 / 2 $$

Short answer

Complete the sentence with the correct answer: The value of \(y_0\) is _______, the value of \(y_0'\) is \(-\sqrt{3}\pi\), while the values of constants a and b cannot be uniquely determined.

Step-by-step solution

Find the first derivative of the given function

To find the first derivative of y(t), we will differentiate the given function with respect to t: $$ y(t) = \sqrt{3}\cos{\pi t} - \sin{\pi t} $$ $$ y'(t) = -\sqrt{3}\pi\sin{\pi t} - \pi\cos{\pi t} $$

Find the second derivative of the given function

To find the second derivative of y(t), we will differentiate the first derivative with respect to t: $$ y'(t) = -\sqrt{3}\pi\sin{\pi t} - \pi\cos{\pi t} $$ $$ y''(t) = -\sqrt{3}\pi^2\cos{\pi t} + \pi^2\sin{\pi t} $$

Evaluate y(t), y'(t), and y''(t) at t = t₀

We will now find the values of y(t), y'(t), and y''(t) at t = 1/2: $$ y\left(\frac{1}{2}\right) = \sqrt{3}\cos{\frac{\pi}{2}} - \sin{\frac{\pi}{2}} = 0 - 1 = -1 $$ $$ y'\left(\frac{1}{2}\right) = -\sqrt{3}\pi\sin{\frac{\pi}{2}} - \pi\cos{\frac{\pi}{2}} = -\sqrt{3}\pi - 0 = -\sqrt{3}\pi $$ $$ y''\left(\frac{1}{2}\right) = -\sqrt{3}\pi^2\cos{\frac{\pi}{2}} + \pi^2\sin{\frac{\pi}{2}} = 0 + \pi^2 = \pi^2 $$

Determine the constants a, b, y₀, and y₀'

Using the obtained values, we will determine the constants: Since y(t₀) = y₀, we have: $$ y_0 = -1 $$ Since y'(t₀) = y₀', we have: $$ y_0' = -\sqrt{3}\pi $$ Finally, using the equation y''(t) + a y'(t) + b y(t) = 0 and the values of y''(1/2), y'(1/2), and y(1/2), we obtain: $$ \pi^2 - a\sqrt{3}\pi - b(-1) = 0 $$ However, we cannot determine the unique values for a, and b from this equation. Therefore, the constants a and b cannot be uniquely determined from the given information. In conclusion: $$ y_0 = -1 $$ $$ y_0' = -\sqrt{3}\pi $$ The constants a and b cannot be uniquely determined.

Key concepts

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. Fundamentally, they are used to model how a particular quantity changes over time or space. For the problem at hand, we have a second order differential equation given by \( y^{\prime \prime} + ay^{\prime} + by = 0 \).
This equation indicates that the acceleration component \( y^{\prime \prime} \) of the function \( y(t) \) is related to its velocity \( y^{\prime} \) and position \( y \) by some constants \( a \) and \( b \). The skill in solving such equations lies in understanding both the mathematical techniques required and the contextual implications they have on the system being modeled. For instance, in physics, such equations can characterize the motion of a damped harmonic oscillator.

Boundary Value Problems

Boundary value problems are a type of differential equation problem where the solution is determined based on the values of the function at specific points, which are called 'boundaries.' However, in this particular problem, it's more accurate to talk about initial conditions rather than boundaries. We're dealing with an initial value problem (IVP) because we know the value of the function and its derivative at a particular point in time, \( t_0 \).
IVPs are common in real-life scenarios where a system's state is known at a starting point, and we seek to understand its behavior at subsequent times. The presence of the initial conditions, \( y(t_0) = y_0 \), and \( y^{\prime}(t_0) = y_0^{\prime} \), allows us to solve for specific solutions that satisfy both the differential equation and the initial conditions.

Solution of Differential Equations

The solution of a differential equation is a function that satisfies the equation for all values within a certain domain. Usually, finding a solution involves integrating the differential equation. In the context of our given problem, the process involved differentiating the provided function to obtain both the first and second derivatives. These derivatives were evaluated at the initial point \( t_0 \) to find the constants associated with the initial conditions.
However, it's crucial to note that without further information or more constraints, the values of \( a \) and \( b \) in this equation cannot be uniquely determined. This demonstrates that some problems may have multiple solutions or even require additional information to find a unique solution. When approaching similar problems, students should ensure they have all the necessary information to find a conclusive answer, and if not, they should describe the set of possible solutions just as we acknowledge the unknown constants \( a \) and \( b \) in the original exercise.