Question

(a) Verify that the given function, \(y_{P}(t)\), is a particular solution of the differential equation. (b) Determine the complementary solution, \(y_{C}(t)\). (c) Form the general solution and impose the initial conditions to obtain the unique solution of the initial value problem. $$y^{\prime \prime}+y=2 t-3 \cos 2 t, \quad y(0)=0, \quad y^{\prime}(0)=0, \quad y_{P}(t)=2 t+\cos 2 t$$

Short answer

Based on the solution above, create a short answer: To solve the initial value problem with the given differential equation, \(y'' + y = 2t - 3\cos{2t}\), and initial conditions, \(y(0) = 0\) and \(y'(0) = 0\), we first verified the proposed particular solution \(y_P(t) = 2t + \cos{2t}\). We then found the complementary solution \(y_C(t) = C_1\cos{t} + C_2\sin{t}\), and combined the complementary and particular solutions to form the general solution: \(y(t) = C_1\cos{t} + C_2\sin{t} + 2t + \cos{2t}\). After applying the initial conditions, we determined the unique solution of the initial value problem to be \(y(t) = -\cos{t} - 2\sin{t} + 2t + \cos{2t}\).

Step-by-step solution

Verification of the given function, \(y_P(t)\)

First, we need to find the first and second derivatives of \(y_P(t)\) in order to verify if it's a particular solution of the given differential equation. Take the first derivative: $$y_P'(t) = \frac{d}{dt}(2t + \cos{2t}) = 2 - 2\sin{2t}$$ Take the second derivative: $$y_P''(t) = \frac{d^2}{dt^2}(2t + \cos{2t}) = -4\cos{2t}$$ Now substitute \(y_P(t)\), \(y_P'(t)\), and \(y_P''(t)\) into the given differential equation: $$(-4\cos{2t}) + (2t+\cos{2t}) = 2t - 3\cos{2t}$$ Since the left side is equal to the right side, \(y_P(t)\) is a particular solution of the given differential equation.

Find the complementary solution, \(y_C(t)\)

To find the complementary solution, first, consider the homogeneous equation: $$y'' + y = 0$$ This is a second-order linear homogeneous differential equation. The characteristic equation is: $$r^2 + 1 = 0$$ Solving for \(r\), we get: $$r = \pm i$$ Hence, the complementary solution is: $$y_C(t) = C_1\cos{t} + C_2\sin{t}$$

Form the general solution

The general solution is the sum of the particular and complementary solutions: $$y(t) = y_C(t) + y_P(t) = C_1\cos{t} + C_2\sin{t} + 2t + \cos{2t}$$

Impose the initial conditions to obtain the unique solution

The initial conditions are: $$y(0) = 0, \quad y'(0) = 0$$ Apply the initial conditions to \(y(t)\) and its derivative: $$y(0) = C_1\cos{0} + C_2\sin{0} + 2(0) + \cos{2(0)} = C_1 + 1 = 0$$ $$y'(0) = -C_1\sin{0} + C_2\cos{0} + 2 + (-2\sin{2(0)}) = C_2 + 2 = 0$$ Solving for \(C_1\) and \(C_2\), we get: $$C_1 = -1, \quad C_2 = -2$$

Obtain the unique solution of the initial value problem

Using the values of \(C_1\) and \(C_2\), and the general solution, we find the unique solution of the initial value problem: $$y(t) = -\cos{t} -2\sin{t} + 2t + \cos{2t}$$

Key concepts

Complementary Solution

When dealing with second-order linear differential equations, understanding the concept of a complementary solution is crucial. A complementary solution, often denoted as \(y_C(t)\), refers to the general solution of the corresponding homogeneous differential equation, which is obtained by setting the non-homogeneous term (if any) to zero. In simpler terms, it's the part of the solution that accounts for the behavior of the equation without any external influences.
For instance, if our non-homogeneous differential equation is \(y'' + y = f(t)\), where \(f(t)\) is a given function, then the corresponding homogeneous equation would be \(y'' + y = 0\). To find \(y_C(t)\), we solve this homogeneous equation using characteristic equations or other suitable methods for solving second-order linear differential equations.

Characteristic Equation and Its Roots

We form a characteristic equation from the homogeneous equation by assuming a solution of the form \(e^{rt}\). In the provided exercise, the characteristic equation is \(r^2 + 1 = 0\), which gives us complex roots \(r = \pm i\). These roots tell us that the complementary solution will consist of sine and cosine functions - which are typical solutions for characteristic equations with complex roots.
In summary, the complementary solution is vital to capture the essence of the differential equation's internal behavior, and for our example exercise, the complementary solution is \(y_C(t) = C_1\cos{t} + C_2\sin{t}\).

Initial Value Problem

An initial value problem is a type of differential equation that comes with specific conditions, known as initial conditions. These conditions are the values of the unknown function (and possibly its derivatives) at a certain point, typically when \(t = 0\). Initial value problems are significant in that they allow us to find a unique solution that fits the particular scenario defined by those initial conditions.
In our exercise, the initial conditions are given as \(y(0) = 0\) and \(y'(0) = 0\). Such conditions are essential both to the physical interpretation of the problem and to guarantee the uniqueness of the solution in a mathematical context. They determine the constants \(C_1\) and \(C_2\) in the complementary solution, which makes the general solution cater to the specific problem at hand.

Applying the Initial Conditions

To apply these initial conditions to our general solution, we substitute \(t = 0\) and equate the resulting expressions to the values given by the conditions. This allows us to solve for \(C_1\) and \(C_2\), leading us to the unique solution that not only satisfies the differential equation but also fits the particular context defined by the initial conditions. This process is evident in steps 4 and 5 of our exercise's solution.

Second-order Linear Differential Equation

A second-order linear differential equation is characterized by the present of the unknown function and its first two derivatives, typically formulated as \(ay'' + by' + cy = f(t)\), where \(a\), \(b\), and \(c\) are coefficients, and \(f(t)\) is a function of \(t\) or a constant. The order of the equation, in this case, second-order, refers to the highest derivative present in the equation.
This type of equation arises frequently in the study of physical systems such as mechanical vibrations, electrical circuits, and dynamic systems. The solutions to these equations give insight into the behavior of such systems over time.

Homogeneous vs. Non-Homogeneous Equations

These differential equations can be either homogeneous, \(f(t) = 0\), or non-homogeneous (otherwise). As seen in our exercise, the solution to a non-homogeneous second-order linear differential equation is comprised of two parts: the complementary solution \(y_C(t)\) and a particular solution \(y_P(t)\). The particular solution is a specific solution that satisfies the non-homogeneous equation.
By combining the complementary solution and the particular solution, we get the general solution to the non-homogeneous differential equation - a crucial concept as it represents all potential behaviors of the system described by the equation. The exercise provided shows a step by step approach to finding such a general solution, making it an excellent example of how to handle second-order linear differential equations.