Question

In each exercise, the unique solution of the boundary value problem is given. Determine the constants \(\alpha, \beta\), and \(\gamma\). $$ y^{\prime \prime}+\gamma y=0, \quad y^{\prime}(0)=\alpha, \quad y(1)=\beta $$

Short answer

Question: Determine the constants α, β, and γ for the unique solution of the boundary value problem y(t) = 2t - 1. Answer: The constants are α = 2, β = 1, and γ = 0.

Step-by-step solution

Differentiate y(t) once and twice

To find y'(t) and y''(t), we need to differentiate y(t) with respect to t: \(y(t) = 2t - 1\) First derivative: \(y'(t) = \frac{d(2t - 1)}{dt} = 2\) Second derivative: \(y''(t) = \frac{d^2(2t - 1)}{dt^2} = 0\) Now we have y'(t) = 2 and y''(t) = 0.

Use first boundary condition to find α

Given y'(0) = α. So, substitute t=0 into y'(t) to find α: \(y'(0) = 2\) Therefore, α = 2.

Use second boundary condition to find β

Given y(1) = β. So, Substitute t=1 into y(t) to find β: \(y(1) = 2(1) - 1 = 1\) Therefore, β = 1.

Find γ using y''(t) and boundary value problem equation

We are given the boundary value problem equation: \(y'' + \gamma y = 0\) Since we found y''(t) = 0 in step 1, we can substitute it into the equation: \(0 + \gamma y = 0\) Now, we can use either y'(t) or y(t) to find γ. Lets use y(t), as it is the original given solution: \(0 + \gamma(2t - 1) = 0\) Since this equation will hold for all values of t, the only possible solution for this is that γ = 0.

Conclusion

We have found all the required constants: α = 2 β = 1 γ = 0

Key concepts

Differential Equations

Differential equations are a cornerstone of mathematics and are key in understanding changes within systems such as physics, engineering, and other scientific disciplines. At their most fundamental, differential equations are mathematical equations that relate a function with its derivatives.

When it comes to solving a differential equation, one crucial endeavor is to find a specific function—or solution—that satisfies the equation. The solution to a differential equation represents the rule that dictates how a particular quantity changes over time or another variable. There are various types of differential equations, but one commonality among them is the use of boundary value conditions or initial conditions to find unique solutions.

In the exercise given, the differential equation takes the form of a second-order linear homogeneous equation:
\( y'' + \(\gamma\) y = 0 \), which intuitively represents a system that has a restoring force proportional to the displacement with a factor \(\gamma\). The solution to this kind of equation can be a sinusoidal function, an exponential function, or, as in the simplified example given, a linear function.

Initial Conditions

When it comes to differential equations, initial conditions provide specific values at a starting point (often denoted as time \(t = 0\) for time-dependent equations) for the function and possibly its derivatives. These conditions are pivotal because they allow us to narrow down the infinite set of possible solutions to the differential equation to just one unique solution that fits the context of the problem.

In our exercise, we focus on two sorts of conditions: The value of the first derivative \(y'(0) = \(\alpha\)\) is an initial condition, and \(y(1) = \(\beta\)\) is a boundary condition since it provides information at a point other than the initial time. By applying these conditions to our differential equation, we can determine the constants that define the unique solution.

This approach embodies the principle that in physical systems, it's the initial state or boundary configuration that dictates the system's evolution over time. It's especially true in many real-world applications such as vibrating systems, thermal conduction, and fluid dynamics.

Second Derivative

The second derivative is an extension of the concept of the first derivative, but instead of measuring the rate of change, it measures the rate at which the rate of change itself changes. This feature helps in understanding the concavity and inflection points of a function, providing insights into the behavior of its graph, and can reveal the presence of acceleration in physical systems.

In differential equations, the presence of a second derivative, such as \(y''\), can signify that our equation is of second order, which often deals with systems that have an inertial aspect—where not just the position but also the velocity (first derivative) and the acceleration (second derivative) have a role in the system's state.

In the given exercise, after taking the second derivative of the proposed solution \(y(t)\), we find that \(y''(t)\) is equal to zero. This indicates that for the specific solution provided, there's no acceleration component in the system, which aligns with the flat slope of the linear equation \(y(t) = 2t - 1\). Consequently, \(\gamma\), which would typically alter our solution's form, must be zero to satisfy the given differential equation for all values of \(t\).