Question

In Exercises \(1-20\) use the Laplace transform to solve the initial value problem. Where indicated by \(\mathrm{C} / \mathrm{G}\), graph the solution. $$ y^{\prime \prime}+9 y=\left\\{\begin{array}{ll} \cos t, & 0 \leq t<\frac{3 \pi}{2}, \\ \sin t, \quad t \geq \frac{3 \pi}{2} \end{array} \quad y(0)=0, y^{\prime}(0)=0\right. $$

Short answer

Question: Determine the solution y(t) to the given initial value problem: y''(t) + 9y(t) = cos(t) for 0 ≤ t < (3π/2), and y''(t) + 9y(t) = sin(t) for t ≥ (3π/2), with initial conditions y(0) = 0 and y'(0) = 0. Answer: The solution to the given initial value problem is: y(t) = { { (1/8)[cos(3t) - cos(t)], 0 ≤ t < (3π/2) } { (1/8)[sin(3t) - sin(t)], t ≥ (3π/2) } with initial conditions y(0) = 0, and y'(0) = 0.

Step-by-step solution

Compute the Laplace transform of the given differential equation

To compute the Laplace transform of the given equation, we'll apply the L{.} operator on both sides: L{y''(t) + 9y(t)} = L{cos(t) 𝑓𝑜𝑟 0≤t<3π/2, sin(t) 𝑓𝑜𝑟 t≥3π/2} Using the linearity property of Laplace transforms: L{y''(t)} + 9L{y(t)} = L{cos(t)}𝑓𝑜𝑟 0≤t<3π/2, L{sin(t)}𝑓𝑜𝑟 t≥3π/2 Now, we'll compute the Laplace transforms for each function: L{y''(t)} = s^2Y(s) - sy(0) - y'(0) = s^2Y(s) L{y(t)} = Y(s) L{cos(t)} = \frac{s}{s^2+1} L{sin(t)} = \frac{1}{s^2+1} So, the equation becomes: s^2Y(s) + 9Y(s) = \frac{s}{s^2+1} 𝑓𝑜𝑟 0≤t<3π/2, \frac{1}{s^2+1} 𝑓𝑜𝑟 t≥3π/2

Solve for Y(s)

Now we'll solve for Y(s) in both cases: For 0≤t<3π/2: (s^2 + 9)Y(s) = \frac{s}{s^2+1} Y(s) = \frac{s}{(s^2+1)(s^2+9)} For ≥3π/2: (s^2 + 9)Y(s) = \frac{1}{s^2+1} Y(s) = \frac{1}{(s^2+1)(s^2+9)}

Find the inverse Laplace transform of Y(s)

Now we'll find the inverse Laplace transform of Y(s) for both cases: For 0≤t<3π/2: y(t) = L^{-1}{\frac{s}{(s^2+1)(s^2+9)}} = \frac{1}{8}[cos(3t) - cos(t)] For t≥3π/2: y(t) = L^{-1}{\frac{1}{(s^2+1)(s^2+9)}} = \frac{1}{8}[sin(3t) - sin(t)]

Write the solution y(t)

Finally, we can write the solution for y(t) as: y(t) = \left\\{\begin{array}{ll} \frac{1}{8}[cos(3t) - cos(t)], & 0 \leq t<\frac{3 \pi}{2}, \\\ \frac{1}{8}[sin(3t) - sin(t)], \quad t \geq \frac{3 \pi}{2} \end{array} \quad y(0)=0, y^{\prime}(0)=0\right. And this is the solution for the given initial value problem.

Key concepts

Understanding Differential Equations

A differential equation is a mathematical equation that involves the derivatives of a function. These equations are key in describing various physical phenomena, such as motion, heat, and wave propagation. In this exercise, we deal with a second-order differential equation because it contains a second derivative \(y''\). The general form of the equation is \(y'' + 9y = f(t)\), where \(f(t)\) changes based on time \(t\) and includes pieces like \(\cos(t)\) and \(\sin(t)\). Solving differential equations often involves finding a function \(y(t)\) that satisfies the given mathematical conditions.

Exploring Initial Value Problems

An initial value problem (IVP) is a type of differential equation that comes with specific conditions, called "initial conditions." These conditions are provided at the start (initial point) of the analysis. In this problem, the initial conditions are given as \(y(0) = 0\) and \(y'(0) = 0\). Initial value problems are crucial as they allow us to find a unique solution to the differential equation that satisfies these initial conditions.
Initial value problems appear frequently in real-world scenarios where a process starts from a known state and evolves over time, such as the cooling of an object or the discharge of a capacitor in an electrical circuit. By solving the IVP, we ensure that our solution captures the whole dynamics from the initial moment.

Inverse Laplace Transform and Its Use

The Laplace transform is a powerful tool for solving differential equations, especially when combined with initial value problems. It transforms complex differential equations into simpler algebraic equations. This transformation can be reversed to find the solution in the time domain — a process known as the inverse Laplace transform.
In this exercise, after applying the Laplace transform on the differential equation and solving for \(Y(s)\), we find its inverse Laplace transform to obtain \(y(t)\). This allows us to write the solution as a real-time function that matches the behavior defined by the original differential equation and initial conditions.

Working with Piecewise Functions

Piecewise functions are functions defined by different expressions based on the input value or domain section. In our problem, the function \(f(t)\) switches from \(\cos(t)\) to \(\sin(t)\) at a critical point : \(t = \frac{3\pi}{2}\).
Tackling piecewise functions involves treating each section separately, then ensuring they connect smoothly at the transition point, like solving our differential equation for two separate cases: \(0 \leq t < \frac{3\pi}{2}\) and \(t \geq \frac{3\pi}{2}\).
This approach is essential for modeling processes that change behavior under different conditions, such as a vehicle switching gears or an electronic signal alternating frequency.