Question

For the given differential equation, (a) Determine the roots of the characteristic equation. (b) Obtain the general solution as a linear combination of real-valued solutions. (c) Impose the initial conditions and solve the initial value problem. $$ y-2 \sqrt{2} y+3 y=0, \quad y(0)=-1 / 2, \quad y(0)=\sqrt{2} $$

Short answer

Question: Determine the general solution of the nonhomogeneous second-order linear differential equation given by \(y''(t) - y'(t) = (t - 1)e^t\), with \(y_1(t) = t\) and \(y_2(t) = e^t\) as solutions of the homogeneous equation. Answer: The general solution of the nonhomogeneous second-order linear differential equation is given by \(y(t) = (c_1t + c_2e^t) + (t^2 + e^t(t-1))\), where \(c_1\) and \(c_2\) are constants.

Step-by-step solution

Complementary Solution

We are given that both \(y_1(t) = t\) and \(y_2(t) = e^t\) are solutions of the homogeneous equation. Therefore, the complementary solution is given by a linear combination of these two functions: $$ y_c(t) = c_1y_1(t) + c_2y_2(t) = c_1t + c_2e^t. $$

Particular Solution using Variation of Parameters

We will now find a particular solution to the nonhomogeneous equation using the method of variation of parameters. The Wronskian of \(y_1(t)\) and \(y_2(t)\) is given by: $$ W(y_1,y_2) = \begin{vmatrix} t & e^t \\ 1 & e^t \end{vmatrix} = te^t - e^t. $$ Next, we compute the particular solution \(y_p(t)\) using the given nonhomogeneous term \((t-1)e^t\) and the Wronskian found above: $$ y_p(t) = -y_1(t)\int\frac{y_2(t)}{W(y_1,y_2)} G(t) \,dt + y_2(t)\int\frac{y_1(t)}{W(y_1,y_2)} G(t) \,dt, $$ where \(G(t) = (t-1)e^t\). Substituting the values, we get: $$ y_p(t) = -t\int\frac{e^t}{te^t - e^t} (t-1)e^t \,dt + e^t\int\frac{t}{te^t - e^t} (t-1)e^t \,dt. $$ Now, we simplify and integrate the two terms: $$ y_p(t) = -t\int\frac{(t-1)}{t-1}dt + e^t\int\frac{t(t-1)}{t(t-1)} dt. $$ After integration, we obtain: $$ y_p(t) = t^2 + e^t(t-1). $$

General Solution

Finally, the general solution of the nonhomogeneous differential equation is the sum of the complementary solution and the particular solution: $$ y(t) = y_c(t) + y_p(t) = (c_1t + c_2e^t) + (t^2 + e^t(t-1)). $$

Key concepts

Characteristic Equation

The characteristic equation is a fundamental concept in the study of differential equations, providing the key to understanding the behavior of solutions to linear homogeneous differential equations. It is crafted by transforming a differential equation into an algebraic equation by assuming the solution will be exponential in nature, generally represented as e^rt, where r is the root of the characteristic equation.

In any linear homogeneous differential equation with constant coefficients, each term of the solution is tied to a root of the characteristic equation. For example, if our characteristic equation is r² - ar - b = 0, its roots will define the nature of the solution, whether it be real distinct, complex, or repeated roots. Determining the roots allows us to construct the complementary solution to the differential equation, which in our practice problem, serves as the backbone of the general solution.

General Solution of Differential Equations

The general solution of a differential equation combines the complementary (homogeneous) and particular (inhomogeneous) solutions. This all-encompassing solution is capable of satisfying any boundary or initial conditions applied to the differential equation. In practice, the complementary solution, y_c(t), is obtained from the characteristic equation of the associated homogeneous problem and is composed of a linear combination of functions that are dependent on the nature of the roots.

For example, if the roots are real and distinct, the complementary solution will be a combination of exponential functions. If they are complex, it will involve sine and cosine terms. In the exercise, our complementary solution takes the form c₁t + c₂e^t. The particular solution, y_p(t), needs to be added to account for non-homogeneous terms and is typically found using the method of undetermined coefficients or variation of parameters. Therefore, the general solution is the sum of these two solutions and holds the form y(t) = y_c(t) + y_p(t).

Initial Value Problem

An initial value problem is a differential equation accompanied by specific conditions that the solution must satisfy at the beginning of the interval of interest, often referenced as the initial conditions. These conditions provide the exact values required to determine the unknown constants in the general solution, making the solution unique to that particular problem.

To solve an initial value problem, we first find the general solution of the differential equation and then apply the initial conditions to solve for the arbitrary constants. If our problem has an initial condition y(0) = y₀, we would substitute t=0 and y=y₀ into our general solution and solve for the constants. The exercise provided involves this step, aiming to find specific values for c₁ and c₂ by using the initial values of the function and its derivative at t=0.

Method of Variation of Parameters

The method of variation of parameters is an advanced technique for finding particular solutions to nonhomogeneous linear differential equations. Unlike to the method of undetermined coefficients, which works well for specific types of right-hand side functions, variation of parameters is a more universal approach that can be applied regardless of the form of the nonhomogeneous term.

The essence of this method involves the assumption that the coefficients of the complementary solution, normally constants, can vary with time. By introducing this flexibility, we are able to account for the particular structure of the inhomogeneous term. The process entails finding the Wronskian determinant of the fundamental set of solutions and integrating with respect to the nonhomogeneous functions. This method is exemplified in the step-by-step solution given for our exercise, where the particular solution, y_p(t), is constructed by integrating with functions derived from the original nonhomogeneous differential equation.