Question

In each exercise, the general solution of the differential equation \(y^{\prime}+p(t) y=g(t)\) is given, where \(C\) is an arbitrary constant. Determine the functions \(p(t)\) and \(g(t)\). \(y(t)=C e^{-2 t}+t+1\)

Short answer

Question: Determine the functions \(p(t)\) and \(g(t)\) in the following differential equation, knowing its general solution: \(y'(t) + p(t)y(t) = g(t)\); General solution: \(y(t) = Ce^{-2t} + t + 1\). Answer: The functions \(p(t)\) and \(g(t)\) in the given differential equation are \(p(t) = 2\) and \(g(t) = 2t + 3\).

Step-by-step solution

Compute the derivative of y(t)

The first step is to compute the derivative of the given function: \(y(t)=Ce^{-2t} + t + 1\) Applying the rules of derivatives we get: \(y'(t)=-2Ce^{-2t} + 1\)

Plug y(t) and y'(t) into the differential equation

Let's plug both \(y(t)\) and its derivative \(y'(t)\) into the differential equation: \(y'(t) + p(t)y(t) = g(t)\) Insert the expressions for \(y(t)\) and \(y'(t)\) into the equation: \((-2Ce^{-2t} + 1) + p(t)(Ce^{-2t} + t + 1) = g(t)\)

Separate the terms with and without C

In order to find \(p(t)\) and \(g(t)\), we want to separate the terms involving \(C\) and the terms which do not involve \(C\). So the equation becomes: \((-2Ce^{-2t} + p(t)Ce^{-2t}) + (p(t)t + p(t) + 1) = g(t)\) Now, we have: \(C(-2e^{-2t} + p(t)e^{-2t}) + p(t)t + p(t) + 1 = g(t)\)

Analyze the coefficients and determine p(t) and g(t)

We can now determine \(p(t)\) and \(g(t)\) by looking at the coefficients of \(C\) and the remaining terms. Considering the terms involving C: \(-2e^{-2t} + p(t)e^{-2t} \implies p(t)e^{-2t} - 2e^{-2t}\) In order for these terms to cancel out (as we are not given any space for a new constant in the general solution), the following should hold: \(p(t)e^{-2t} - 2e^{-2t} = 0\) Hence, \(p(t)e^{-2t} = 2e^{-2t} \implies p(t) = 2\) Now considering the remaining terms: \(g(t) = p(t)t + p(t) + 1\) Reemplacing \(p(t) = 2\), we find: \(g(t) = 2t + 2 + 1\) So, \(g(t) = 2t + 3\) The functions \(p(t)\) and \(g(t)\) are, respectively: \(p(t) = 2\) \(g(t) = 2t + 3\)

Key concepts

General Solution

The term 'general solution' in the context of differential equations refers to a solution that contains all possible particular solutions of a differential equation. It incorporates arbitrary constants due to the integration process involved in solving the equation.

For example, in the exercise provided, the general solution of the differential equation is given as
\(y(t)=Ce^{-2t} + t + 1\), where \(C\) is the arbitrary constant. This general solution is effectively a family of solutions depending on different values of \(C\), each corresponding to a unique particular solution fitting to specific initial conditions. Understanding the concept of general solution is essential because it shows the whole set of possible behaviors that the system described by the differential equation can exhibit.

When working back from a general solution to find the particular functions in a differential equation, like in the exercise, it is crucial to identify terms involving the arbitrary constant and those who do not to separate the effects of the integrating factor and the non-homogeneous part of the solution.

Integrating Factor

An 'integrating factor' is a function that is used to simplify the solving process of a first-order linear differential equation. The idea is to multiply the entire differential equation by an integrating factor so that the left-hand side becomes the derivative of a product of two functions, which makes it easier to integrate and find the general solution.

While the original exercise did not require calculating an integrating factor as it provided the general solution, it is essential to recognize that the integrating factor method is implied in obtaining that solution. The presence of the exponential term \(e^{-2t}\) in the general solution often hints at the integrating factor that would have been used if we were solving the differential equation from scratch.

It's noteworthy that in the differential equation \(y'+p(t)y=g(t)\), if one had to solve for \(y(t)\), the integrating factor \(\mu(t)\) would typically be found using the formula \(\mu(t) = e^{\int p(t)dt}\). This knowledge enhances comprehension of how the particular form of the general solution was likely achieved.

First Order Differential Equations

First order differential equations are characterized by the highest derivative being the first derivative of the unknown function. They come in various forms, including linear, separable, and exact equations, among others. The exercise at hand deals with a first-order linear differential equation of the form
\(y^{\'prime} + p(t) y = g(t)\).

The challenge in solving first-order differential equations lies in identifying the method that suits the equation's structure. When the equation is linear and of the form provided, it is possible to express its solution via the superposition principle as a homogeneous solution plus a particular solution. The homogeneous solution involves the arbitrary constant \(C\) and the given integrating factor, whereas the particular solution is usually found by methods like undetermined coefficients or variation of parameters.

The ability to recognize and classify first order differential equations allows students to employ the correct strategy, such as the integrating factor method demonstrated indirectly in the solution of the exercise, to find the general solution effectively.