Question

For each initial value problem, (a) Find the general solution of the differential equation. (b) Impose the initial condition to obtain the solution of the initial value problem. \(y^{\prime}+y=1+2 e^{-t} \cos 2 t, \quad y(\pi / 2)=0\)

Short answer

Based on the solution, create a short answer question: Question: Determine the integrating factor of the given first-order linear ordinary differential equation with a non-homogeneous term: \(y'+y=e^t+2e^{-t}\cos{2t}\). Answer: The integrating factor is \(e^{-t}\).

Step-by-step solution

Find the integrating factor

Let's first find the integrating factor (IF). For this equation, the IF is given by \(e^{\int P(t) dt}\), where \(P(t)=-1\). The integral is \(\int (-1) dt = -t + C\). So, the integrating factor is \(e^{-t}\).

Multiply the equation by the integrating factor and integrate

Now, we will multiply the given equation by the integrating factor (IF) and rewrite the equation as \((ye^{-t})'=1e^{-t}+2e^{-2t}\cos{2t}\). Integrate both sides of the equation with respect to \(t\): \(\int(ye^{-t})' dt = \int(1e^{-t} + 2e^{-2t}\cos{2t}) dt\) After integration, we get: \(ye^{-t} = -e^{-t} + \int 2e^{-2t}\cos{2t} dt\)

Find the constant of integration using the initial condition

To find the constant of integration we use the initial condition provided in the exercise, which is \(y(\frac{\pi}{2})=0\). When substituting \(t=\frac{\pi}{2}\), we get: \(0 = -e^{-\frac{\pi}{2}} + \int 2e^{-\pi}\cos{\pi} dt\) This implies that, \(\int 2e^{-\pi}\cos{\pi} dt = e^{-\frac{\pi}{2}}\) Unfortunately, the integral part cannot be integrated using elementary functions. The result is in terms of special functions. So, we'll leave the integral as it is and rewrite the equation as: \(ye^{-t} = -e^{-t} + e^{-\frac{\pi}{2}} +\int 2e^{-2t}\cos{2t} dt\)

Write down the final solution

Finally, we will multiply both sides of the equation by \(e^t\) to obtain the solution for the initial value problem: \(y(t) = -1 + e^{\frac{\pi}{2}-t} + e^t\int 2e^{-2t}\cos{2t} dt\) As integral part cannot be simplified further, this is the final solution to the given initial value problem.

Key concepts

Differential Equations

Differential equations are mathematical models that describe the relationship between a function and its derivatives. They are used to model various phenomena such as growth rates, motion, and changes in temperature, to name a few. A differential equation entails an unknown function, often denoted by y(t) or f(x), and one or more of its derivatives.For instance, the equation from the exercise, \(y^{\'RR'}+y=1+2 e^{-t} \times \times 2 t\), is a first-order linear differential equation. It says that the rate of change of the function y(t) is related to y(t) itself and a function of t.In general, differential equations are categorized by order, denoted by the degree of the highest derivative present, and by linearity, determined by whether or not the equation involves the function and its derivatives to the first power only (no y^2 or y'^2, for example). The given exercise involves a first-order linear differential equation because it has the first derivative of y, y', and the terms of y and y' are to the first power.

Integrating Factor Method

The integrating factor method is a technique used to solve certain types of differential equations, namely linear differential equations. The main idea is to multiply the equation by an intelligently chosen function, called an integrating factor (IF), which simplifies the equation to a form that can be integrated directly.The integrating factor is usually a function of the independent variable (in the given exercise, t), and it is chosen so that the left side of the equation, once multiplied by the IF, becomes the derivative of a product of functions. In our example, the IF is determined by \(e^{\int P(t) dt}\), where P(t) is the coefficient of y in the differential equation. After identifying the IF, the original equation is multiplied by this factor, and both sides are integrated to find a general solution.This method is elegant and powerful because it transforms a differential equation into a simpler form, allowing us to use straightforward integration to progress towards the solution.

General Solution of Differential Equations

The general solution of a differential equation represents a family of functions that comprises all possible solutions to the equation. In essence, it includes an arbitrary constant (or constants) that can be adjusted to meet initial or boundary conditions. This is in contrast to a particular solution, which is a specific member of the family that satisfies the given conditions of the problem.In the process of solving a differential equation, we often integrate one or more times, introducing constants of integration. These constants are what provide us with a general solution to the differential equation. The general solution encapsulates countless potential functions, where each function corresponds to a different value of the constants. The exercise provided involves finding such a general solution before utilizing the given initial condition to pin down the exact, specific solution that aligns with that condition.

Boundary Conditions

Boundary conditions are constraints necessary to determine a unique solution to a differential equation from its general solution. They are values of the function or its derivatives at specific points, typically the 'boundaries' of the domain in question. Initial conditions are a type of boundary conditions where the values are known at a particular starting point.When solving an initial value problem, as in our exercise, the solution we find from the differential equation must also satisfy the initial condition given. In practical terms, we take the general solution we've discovered, apply the boundary conditions (in this case, initial conditions), and solve for the constants of integration. After imposing the initial condition \(y(\pi / 2)=0\), we are left with a specific equation describing the behavior of y(t) that is consistent not only with the differential equation but also with the initial state of the system we're modeling.