Question

Consider the initial value problem \(y^{\prime \prime}+\omega^{2} y=g(t), y(0)=0, y^{\prime}(0)=0\), where \(\omega\) is a real nonnegative constant. For the given function \(g(t)\), determine the values of \(\omega\), if any, for which the solution satisfies the constraint \(|y(t)| \leq 2,0 \leq t<\infty\). $$ g(t)=\sin \omega t $$

Short answer

IVP: \(y^{\prime \prime}+\omega^{2} y=\sin(\omega t)\), with \(y(0)=0\), \(y^{\prime}(0)=0\), and \(|y(t)| \leq 2\) for \(0 \leq t<\infty\). Answer: The solution satisfies the constraint for \(\omega \ge \frac{1}{\sqrt{2}}\).

Step-by-step solution

1. Find the particular solution \(Y_p(t)\)

To find the particular solution, we first assume a trial solution of the form \(Y_p(t)=A\sin(\omega t) + B\cos(\omega t)\). Let's find the first and second derivatives of \(Y_p(t)\): $$ Y_p^\prime(t) = A\omega\cos(\omega t) - B\omega\sin(\omega t) $$ and $$ Y_p^{\prime\prime}(t) = -A\omega^2\sin(\omega t) - B\omega^2\cos(\omega t) $$ Now, substitute \(Y_p^{\prime\prime}(t)\) and \(Y_p(t)\) into the nonhomogeneous equation and get: $$ -A\omega^2\sin(\omega t) - B\omega^2\cos(\omega t) + \omega^2(A\sin(\omega t) + B\cos(\omega t)) = \sin(\omega t). $$ Simplify this expression, and we get: $$ (A-B)\omega^2\sin(\omega t) + (A+B)\omega^2\cos(\omega t) = \sin(\omega t). $$ Separating coefficients, we obtain the following system of equations for \(A\) and \(B\): $$ \begin{cases} (A-B)\omega^2 = 1\\ (A+B)\omega^2 = 0 \end{cases} $$ Solve the second equation for \(B\), $$ B = -A $$ and substitute it into the first equation: $$ (2A)\omega^2 = 1 $$ which gives us $$ A=-B =\frac{1}{2\omega^2}. $$ Thus, the particular solution is $$ Y_p(t)=\frac{1}{2\omega^2}(\sin(\omega t)-\cos(\omega t)). $$

2. Find the complementary solution \(Y_c(t)\)

As the given differential equation is linear homogeneous, its corresponding homogeneous equation is $$ y^{\prime\prime}+\omega^2y=0. $$ The homogeneous equation has the general solution: $$ Y_c(t) = C_1\sin(\omega t) + C_2\cos(\omega t). $$

3. Apply the initial conditions and find the general solution

We now apply the initial conditions \(y(0)=0\) and \(y'(0)=0\) to find the constants \(C_1\) and \(C_2\). From the initial condition \(y(0)=0\), we have $$ Y_c(0) = C_1\sin(0) + C_2\cos(0) = C_2 = 0. $$ Now, using the initial condition \(y'(0)=0\), and the first derivative of the complementary solution, we get $$ Y_c^\prime(0) = \omega C_1\cos(0) + \omega C_2\sin(0) = \omega C_1 = 0. $$ Since \(\omega\) is nonnegative, we have \(C_1=0\). Thus, the complementary solution \(Y_c(t)\) is 0 and the general solution is given by $$ y(t) = Y_c(t) + Y_p(t) = \frac{1}{2\omega^2}(\sin(\omega t) - \cos(\omega t)). $$

4. Check the constraint and find values of ω

We have the constraint \(|y(t)|\leq 2\) for \(0 \leq t<\infty\). By analyzing the general solution, we observe that \(|y(t)|\le \frac{1}{2\omega^2}(1+1) = \frac{1}{\omega^2}\). Thus, to satisfy the constraint, we need: $$ \frac{1}{\omega^2} \leq 2. $$ From this inequality, we obtain $$ \omega^2 \geq \frac{1}{2}. $$ and $$ \omega \geq \frac{1}{\sqrt{2}}. $$ So, for \(\omega \ge \frac{1}{\sqrt{2}}\), the solution satisfies the constraint \(|y(t)|\leq 2\) for \(0\leq t<\infty\).

Key concepts

Differential Equations

Differential equations are mathematical expressions that relate some function with its derivatives. In the context of physical problems, they can represent a variety of phenomena, such as motion, growth, decay, and other time-dependent processes. The order of the differential equation is determined by the highest derivative present in the equation.

An initial value problem involves a differential equation along with conditions specifying the value of the function and its derivatives at a single point. These initial conditions allow us to find one particular solution from the infinite family of possible solutions to a differential equation, making the solution unique.

Particular Solution

A particular solution to a differential equation is a solution that not only satisfies the differential equation but also satisfies given initial conditions or boundary values. In the exercise, the particular solution, denoted by \(Y_p(t)\), is crafted by using a method that accounts for the nonhomogeneous term \(g(t)\), and ensures the resulting expression when plugged into the differential equation yields a true statement. This solution is specific to the nonhomogeneous part of the equation and helps construct the general solution.

Complementary Solution

The complementary solution, \(Y_c(t)\), refers to the general solution of the corresponding homogeneous differential equation, which is the equation with the right-hand side set to zero. The complementary solution encompasses all possible solutions to the homogeneous part of the differential equation and includes arbitrary constants which are typically resolved using initial conditions. In our case, the complementary solution turned out to be zero as a result of applying the initial conditions \(y(0)=0\) and \(y'(0)=0\).

Initial Conditions

Initial conditions are specific values assigned to the function and possibly some of its derivatives at a particular point. These conditions enable us to narrow down the family of solutions to a single, unique solution. With an initial value problem, the solution must satisfy these conditions at the given point. In this problem, the conditions \(y(0)=0\) and \(y'(0)=0\) were critical for determining the constants in the complementary solution and ensuring that the final solution is applicable to the physical context described by the problem.

Homogeneous Equations

Homogeneous equations are differential equations where every term is a function of the dependent variable and its derivatives. Such equations are set equal to zero. The solutions to homogeneous equations are important as they form the complementary solution which can be used to address the structure of the general solution to the overall problem. Homogeneous equations often represent balanced states where external forces or influences are absent.

Nonhomogeneous Equations

In contrast to homogeneous equations, nonhomogeneous equations include a non-zero term which represents an external influence or forcing function. The problem we're considering has such a term, \(g(t)=\sin(t)\), making the equation nonhomogeneous. Finding the particular solution to this kind of equation is more complicated because it involves identifying a function that meets both the nature of this external term and the original differential equation.

Boundary Value Problems

Boundary value problems are similar to initial value problems but instead specify the values of the solution at more than one point, often at the beginning and end of a given interval. However, in our problem constraint, rather than finding exact values at the boundaries, we assessed the solution's behavior across an infinite interval. The constraint \(|y(t)| \leq 2\) is evaluated for all \(t\geq 0\), which dictates a condition on the magnitude of the function rather than its precise values at the endpoints.