Question

The general solution of the nonhomogeneous differential equation \(y^{\prime \prime}+\alpha y^{\prime}+\beta y=g(t)\) is given, where \(c_{1}\) and \(c_{2}\) are arbitrary constants. Determine the constants \(\alpha\) and \(\beta\) and the function \(g(t)\). $$y(t)=c_{1}+c_{2} e^{-t}+t^{2}$$

Short answer

Short Answer: For the given nonhomogeneous differential equation, the values of the unknowns are α = 0, β = 1, and the function g(t) is given by g(t) = 2 + t^2.

Step-by-step solution

Differentiate y(t) with respect to t

In order to find the first and second derivatives of the given function y(t), we will differentiate it with respect to t. Given: $$ y(t) = c_{1} + c_{2}e^{-t} + t^{2} $$ First derivative of y(t) with respect to t: $$ y^{\prime}(t) = 0+c_{2}( -e^{-t}) + 2t $$ Second derivative of y(t) with respect to t: $$ y^{\prime\prime}(t) = c_{2}e^{-t} + 2 $$

Substitute the derivatives into the nonhomogeneous differential equation

Now, we will substitute the derivatives we found in Step 1 into the differential equation \(y^{\prime\prime}+\alpha y^{\prime}+\beta y=g(t)\). Our aim is to find the unknowns α, β and g(t). Differential equation: $$ y^{\prime\prime}(t) + \alpha y^{\prime}(t) + \beta y(t) = g(t)$$ Substituting the derivatives: $$ (c_{2} e^{-t} + 2) + \alpha (-c_{2} e^{-t} + 2t) + \beta (c_1 + c_{2} e^{-t} + t^2) = g(t) $$

Equate coefficients and find unknowns

Let's equate the coefficients of like terms to find the unknowns, claiming that the two sides of the equation are identical. Equate powers of e^(-t): $$ c_{2} - c_{2} \alpha + \beta c_{2} = 0 $$ This implies that $$ \beta = \alpha $$ Equate coefficients of t^2: $$ \beta = 1 $$ Equate coefficients of t: $$ 2\alpha = 0 $$ We can now find that: $$ \alpha = 0 $$ $$ \beta = 1 $$

Determine g(t)

To find g(t), we will substitute the found values of α and β into the equation from Step 2. $$ g(t) = (c_{2} e^{-t} + 2) - 0 + (1)(c_1 + c_{2} e^{-t} + t^2) $$ $$ g(t) = 2 + t^{2} $$

Summarize Results

We have now found the values of α, β, and the function g(t) for the given nonhomogeneous differential equation. $$ \alpha = 0 $$ $$ \beta = 1 $$ $$ g(t) = 2 + t^{2} $$

Key concepts

Differential Equations

A differential equation is a mathematical equation that relates a function with its derivatives. In other words, it involves the calculation of how a particular function changes and based on that, establishes a relationship. These equations are fundamental in mathematics because they can model a wide array of phenomena in science and engineering, from the motion of planets to the spread of diseases. Differential equations are broadly classified into two main types: ordinary differential equations (ODEs) and partial differential equations (PDEs). An ODE involves derivatives with respect to only one variable, whereas a PDE involves derivatives with respect to multiple variables. Within these classifications, differential equations can also be linear or nonlinear, and homogeneous or nonhomogeneous. A nonhomogeneous differential equation, like the one in our exercise, includes a term that is not merely a combination of the function and its derivatives but also an independent function of the variable, denoted as g(t) in the given problem.

Boundary Value Problems

Boundary value problems (BVPs) are a type of differential equation where the solution is determined by the equation itself and some additional constraints, called boundary conditions. These constraints specify the values (or behaviors) of the solution at specific points, which are called the 'boundaries'. BVPs are crucial in the physical sciences because they accurately describe systems that must satisfy conditions at more than one point, like the temperature distribution in a rod at its endpoints, or the motion of a cable fixed at two points. In our exercise, although boundary conditions aren't explicitly given, the arbitrary constants could be determined with appropriate boundaries. In solving BVPs, it's critical not only to solve the differential equation but also to ensure that the solution meets the boundary conditions set by the specific problem at hand.

Differential Equation Solution

The solution to a differential equation represents a set of functions or a unique function that satisfies the equation. In other words, when you substitute this function and its derivatives into the equation, it should hold true for all applicable values of the variable. There are various methods to solve different types of differential equations, such as separation of variables, integrating factor method, or, for more complicated problems, numerical methods. For nonhomogeneous differential equations, the general solution is comprised of two parts: the complementary function (or homogeneous solution) which solves the associated homogeneous equation, and a particular solution which specifically addresses the nonhomogeneity of the equation. In the exercise we are discussing, y(t) consists of both a homogeneous solution represented by the arbitrary constants and a specific selection for a particular solution, which is a polynomial term t^2, to cater to the nonhomogeneous part of the equation. By following the systematic approach of differentiating and substituting into the original equation, one can determine the undetermined coefficients and the nonhomogeneous function g(t), laying down the path to a complete solution.