Question

Find the general solution of the given differential equation. In Problems 11 and \(12 \mathrm{g}\) is an arbitrary continuous function. $$ y^{\prime \prime}+4 y^{\prime}+4 y=t^{-2} e^{-2 t}, \quad t>0 $$

Short answer

Question: Determine the general solution for the second-order linear non-homogeneous differential equation \(y^{\prime\prime} + 4y^{\prime} + 4y = t^{-2}e^{-2t}, \; t>0\). Answer: The general solution of the given differential equation is \(y(t) = C_1 e^{-2t} + C_2 te^{-2t} + \frac{1}{4t^2} e^{-2t}\) where \(C_1\) and \(C_2\) are arbitrary constants.

Step-by-step solution

Find the complementary function of the homogeneous equation

To find the complementary function, let's consider the homogeneous part of the equation: $$ y^{\prime \prime} + 4y^{\prime} + 4y = 0 $$ The characteristic equation is: $$ m^2 + 4m + 4 = 0 $$ which can be factored as: $$ (m + 2)^2 = 0 $$ So, the repeated root is \(m=-2\). Therefore, the complementary function is given by: $$ y_c(t) = C_1 e^{-2t} + C_2 te^{-2t} $$ where \(C_1\) and \(C_2\) are arbitrary constants.

Find a particular solution for the non-homogeneous equation

To find a particular solution for the non-homogeneous equation, we use the method of undetermined coefficients. The forcing term is \(t^{-2} e^{-2t}\), so we assume a solution of the form: $$ y_p(t) = At^{-2}e^{-2t} $$ Now, let's find the first and second derivatives of \(y_p(t)\): $$ y_p^{\prime}(t) = -2At^{-2} e^{-2t} + 2At^{-3} e^{-2t} $$ and $$ y_p^{\prime \prime}(t) = 4At^{-2} e^{-2t} - 12At^{-3} e^{-2t} + 6At^{-4} e^{-2t} $$ Substituting \(y_p(t)\), \(y_p^{\prime}(t)\), and \(y_p^{\prime \prime}(t)\) into the original non-homogeneous differential equation, and factoring the exponential term, we get: $$ (4At^{-2} - 8At^{-3} + 6At^{-4}) = t^{-2} $$ To solve for A, equate the coefficients of \(t^{-2}\): $$ 4A = 1 \implies A = \frac{1}{4} $$ So, the particular solution is given by: $$ y_p(t) = \frac{1}{4t^2} e^{-2t} $$

Find the general solution

Now, let's add the complementary function and the particular solution to form the general solution: $$ y(t) = y_c(t) + y_p(t) = C_1 e^{-2t} + C_2 te^{-2t} + \frac{1}{4t^2} e^{-2t} $$ This is the general solution of the given differential equation.

Key concepts

General Solution of Differential Equation

A differential equation represents a relationship involving a function and its derivatives. The general solution of a differential equation is a family of solutions that includes every possible solution to the differential equation. In our exercise, you learned that the general solution is a sum of the complementary function and a particular solution. The complementary function, denoted as y_c(t), covers the solution to the homogeneous equation, and the particular solution, y_p(t), is designed to address the non-homogeneity, which in this case is caused by the term t^{-2} e^{-2 t}.

The general solution is expressed as:$y(t)\; =\; y\_c(t)\; +\; y\_p(t)$Understanding this formula is key to solving not just the exercise at hand, but any linear second-order differential equation.

Method of Undetermined Coefficients

When tackling non-homogeneous differential equations, one commonly used technique to find a particular solution is the method of undetermined coefficients. This approach involves guessing a form of the particular solution that resembles the nonhomogeneous term. Then, free parameters (the undetermined coefficients) in that form are found by substituting the guess into the differential equation and solving for these parameters.

In the presented exercise, the assumed form was y_p(t) = At^{-2}e^{-2t}, where A is the undetermined coefficient. By deriving the necessary derivatives and plugging them into the differential equation, we find the value of A that makes the equation work. It's a practical method when the nonhomogeneity is a simple function of t, such as polynomials, exponentials, or trigonometric functions.

Characteristic Equation

When faced with a homogeneous linear differential equation with constant coefficients, the characteristic equation paves the way to find the complementary function. This crucial step translates the differential equation into an algebraic equation whose solutions are known as characteristic roots.

To arrive at the characteristic equation, one typically replaces each derivative of the function y(t) with a term involving m, where m^n represents the nth derivative. Hence, for a second-order differential equation like y'' + 4y' + 4y = 0, substituting y = e^{mt} leads us to the characteristic equation m^2 + 4m + 4 = 0. The roots of this equation dictate the form of the complementary function and reveal whether we'll have real, distinct roots or complex or repeated roots.

Complementary Function

The complementary function, often denoted as y_c(t), represents the general solution to the homogeneous part of a linear differential equation. It embodies the behavior of the system described by the differential equation when external forces (the non-homogeneous part) are absent.

In our example, the homogeneous equation is y'' + 4y' + 4y = 0, and its complementary function is found after determining the roots of the characteristic equation. Since we have a repeated root m = -2, the complementary function takes on the form: y_c(t) = C_1 e^{-2t} + C_2 te^{-2t}, where C_1 and C_2 are arbitrary constants that will be determined by the initial conditions of the specific problem. This concept captures all potential behaviors of the system without external forces, making it a cornerstone of the solution process.