Question

In each exercise, the unique solution of the boundary value problem is given. Determine the constants \(\alpha, \beta\), and \(\gamma\). $$ y^{\prime \prime}+\gamma y=2 e^{t}, \quad y(0)=\alpha, \quad y\left(\frac{\pi}{2}\right)=\beta . \quad \text { The solution is } y(t)=e^{t}+\sin t $$

Short answer

Question: Determine the constants \(\alpha\), \(\beta\), and \(\gamma\) in the equation and boundary conditions given the unique solution \(y(t) = e^t + \sin t\). Answer: After substituting the unique solution and finding the first and second derivatives, we find that \(\alpha = 1\), \(\beta = e^{\frac{\pi}{2}} + 1\), and \(\gamma = 1\).

Step-by-step solution

Find the first and second derivatives of y(t)

To use the given solution in the given differential equation, we need first to find the first and second derivatives of y(t) with respect to t. Given: $$ y(t) = e^t + \sin t $$ Using the chain rule, differentiation of exponential function and trigonometric function, we can compute the derivatives as follows: $$ y'(t) = \frac{d}{dt}(e^t + \sin t) = e^t + \cos t $$ and $$ y''(t) = \frac{d^2}{dt^2}(e^t + \sin t) = e^t - \sin t $$

Plug the solution and its derivatives into the differential equation

Next, we substitute the solution and its derivatives into the given differential equation: $$ y''(t) + \gamma y(t) = 2e^t $$ So, we plug in and get: $$ (e^t - \sin t) + \gamma (e^t + \sin t) = 2e^t $$

Solve for the constant \(\gamma\)

In order to find the value of \(\gamma\), we compare the coefficients of the terms on both sides of the equation. The left side contains the coefficients of \(e^t\) and \(\sin t\): \((1+\gamma)e^t + \gamma\sin t\). This has to be equal to the right side, which has \(2e^t\) as the only term. Comparing coefficients for \(e^t\), we get: $$ 1 + \gamma = 2 $$ Solving for \(\gamma\), we find: $$ \gamma = 1 $$

Plug the solution into the boundary conditions

Now, we can use the boundary conditions to find the constants \(\alpha\) and \(\beta\). We are given: $$ y(0) = \alpha $$ and $$ y\left(\frac{\pi}{2}\right) = \beta. $$ Substitute the provided solution \(y(t)\) into these boundary conditions: $$ \alpha = y(0) = e^0 + \sin 0 = 1 $$ and $$ \beta = y\left(\frac{\pi}{2}\right) = e^{\frac{\pi}{2}} + \sin \frac{\pi}{2} = e^{\frac{\pi}{2}} + 1. $$

Report the values of \(\alpha\), \(\beta\), and \(\gamma\)

Finally, we report the values we found for the constants in the exercise: $$ \alpha = 1, \quad \beta = e^{\frac{\pi}{2}} + 1, \quad \text{and } \gamma = 1. $$

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They appear frequently in the sciences and engineering and describe many phenomena like growth and decay, motion, and the flow of electricity. The most basic form is a first-order differential equation, but they can quickly become more complex, involving higher-order derivatives and multiple variables. Solving these equations often requires finding a function that satisfies the given relation between the derivatives, sometimes under specific initial conditions or boundary values, which is exactly what we encounter in the given boundary value problem.

In our exercise example, the differential equation given is a second-order homogeneous linear equation with constant coefficients. This means that the highest derivative is the second derivative and the equation can be written in a standard form where the derivatives of the function are linearly related to the function itself. The challenge often lies in identifying the methods suitable for solving a particular type of differential equation, with one common approach being to propose a solution involving specific types of functions and then verifying and correcting this solution by inserting it into the equation.

Exponential Function

The exponential function is one of the most important functions in mathematics, especially for solving differential equations. It's defined as a function of the form f(x) = e^x, where e is a mathematical constant approximately equal to 2.71828. Exponential functions are characterized by their rate of growth being proportional to their value, making them naturally appear in real-world processes where things grow rapidly, such as populations or investments.

In the context of our problem, the exponential function appears on the right side of the differential equation as well as in the solution. This function's relevance is highlighted when finding the general solution to differential equations since it's often part of the complementary function, a term used in the method of solving linear differential equations. The constants associated with these functions are determined by the initial or boundary conditions of the specific problem.

Trigonometric Function

Trigonometric functions like sin and cos play a pivotal role in mathematics, engineering, and physics. They relate angles to the sides of a right-angled triangle but also describe repetitive phenomena such as waves. In differential equations, these functions are used to represent solutions to problems involving oscillatory behavior like spring motions or electrical circuits.

In our boundary value problem, the trigonometric function sin is a part of the solution to the differential equation, suggesting that the problem's context could be oscillatory in nature. This inclusion makes the solution more complex and necessitates using trigonometric identities and derivatives when working with differential equations. The challenge with trigonometric functions in these equations is ensuring that the solutions match given boundary or initial conditions, as they can significantly alter the resulting graph of the solution.

Initial Conditions

Initial conditions in differential equations are values that are given for the function or its derivatives at a particular point. These conditions are crucial in determining the unique solution to a differential equation when the general solution involves arbitrary constants. By applying these conditions, those constants can be specifically calculated, producing a unique and particular solution to the problem.

In the solution process of our example problem, we used the boundary conditions y(0) = α and y(π/2) = β to find the constants α and β. Boundary conditions like these are similar to initial conditions but are applied at more than one point, often the endpoints of a domain of interest. These conditions ensure that the solution not only satisfies the differential equation but also adheres to certain predefined constraints, whether they stem from physical boundaries, certain known outcomes at specific points, or conservation laws in physics.