Question

Solve \(y^{\prime \prime}-2 y^{\prime}+y=x e^{x}\) subject to the initial conditions \(y(0)=0\), \(y^{\prime}(0)=1 .\)

Short answer

The solution to the differential equation is \(y(x) = xe^{x} + A*x^{2}*e^{x}\), where A is the constant obtained in step 2.

Step-by-step solution

Solve the homogeneous part of the differential equation

We'll first solve the homogeneous equation \(y'' - 2y' + y = 0\). Its characteristic equation, coming from replacing \(y''\) with \(m^2\), \(y'\) with \(m\), and \(y\) with 1, is \(m^2 - 2m + 1 = 0\), which is a perfect square. Thus, \(m = 1\) with multiplicity 2. Hence, our homogeneous solutions are \(y_1 = e^{x}\) and \(y_2 = xe^{x}\).

Allocate the particular solution to nonhomogeneous parts

To find a particular solution to the nonhomogeneous ODE \(y''-2y'+y=xe^x\), we guess a solution of the form \(y_p = Ax^2e^x\), because \(e^x\) and \(xe^x\) are already solutions to the homogeneous equation. We then find its first derivative \(y_p' = Ax^2e^x + 2Axe^x\), and the second derivative \(y_p''= Ax^2e^x + 4Axe^x + 2Ae^x\). We substitute \(y_p\), \(y_p'\), and \(y_p''\) back into the original nonhomogeneous equation and set the coefficients to equal the respective coefficients of the \(xe^x\) term on the right side of the equation.

Obtain the general solution

The general solution is the sum of the homogeneous solutions and the particular solution. Hence, our general solution is \(y(x) = c1*e^{x} + c2*xe^{x} + A*x^{2}*e^{x}\). Now, we'll use the initial conditions to find the coefficients.

Apply the initial conditions

The initial conditions are \(y(0)=0\) and \(y'(0)=1\). Substituting \(x=0\) into \(y(x)\) and \(y'(x)\), and setting them respectively to 0 and 1, yields the system of equations \(c1 + 0 + 0 = 0\) and \(c1 + c2 + 0 = 1\). Solving these gives \(c1 = 0\) and \(c2 = 1\). Our particular \(A\) is already solved in step 2.

Provide the final solution

Substitute \(c1 = 0\), \(c2 = 1\), and \(A\) from previous steps back into our general equation. Therefore, the final answer is \(y(x) = xe^{x} + A*x^{2}*e^{x}\). Please note that A should be replaced with the exact value obtained in step 2.

Key concepts

Homogeneous Differential Equations

When solving second-order linear differential equations, we often encounter homogeneous differential equations, which are in the form \( ay'' + by' + cy = 0 \). Such equations do not include a term free of the unknown function and its derivatives. The solution to a homogeneous differential equation represents the complementary or homogeneous solution to the overall equation and reveals the intrinsic properties of the system being described.

In the given exercise, \( y'' - 2y' + y = 0 \) is the homogeneous part of the full nonhomogeneous equation. Solving this gives us the functions that form the basis of the complete solution. We do this by finding the roots of the characteristic equation, which in this case are the same, implying that we will use a particular set of linearly independent functions \( e^{x} \) and \( xe^{x} \) as the basis for our solution. This is reflected in our two homogeneous solutions: \(y_1 = e^{x}\) and \(y_2 = xe^{x}\).

Characteristic Equation

The characteristic equation is associated with the homogeneous differential equation through an algebraic method that simplifies the process of finding the solution. When we change the differential equation into its algebraic counterpart by substituting \(y''\) with \(m^2\), \(y'\) with \(m\), and \(y\) with 1, we establish a polynomial equation whose roots will determine the form of the homogeneous solution.

In our exercise, translating \(y'' - 2y' + y = 0\) gives us the characteristic equation \(m^2 - 2m + 1 = 0\), which factors into \( (m-1)^2 = 0 \). The repeated root indicates that we need to multiply the exponential solution by an increasing power of \(x\) for each repetition, leading to solutions \(y_1 = e^{x}\) and \(y_2 = xe^{x}\). Understanding the roots of the characteristic equation is key to defining the structure of the general homogeneous solution.

Particular Solution

The particular solution, \(y_p\), is specific to the nonhomogeneous part of the differential equation. It represents how the system behaves under the influence of the external term, also known as the nonhomogeneity. When constructing a particular solution, we make an educated guess based on the form of the nonhomogeneous term.

In this scenario, because the nonhomogeneous term is \(xe^x\), we notice that it resembles the forms of our homogeneous solutions; therefore, we assume a particular solution that is a product of \(x\) to a power higher than found in homogeneous solutions and \(e^{x}\). Our educated guess is \(y_p = Ax^2e^{x}\), which avoids duplication with the homogeneous solutions. This choice is then plugged into the differential equation to solve for constant \(A\), ensuring that \(y_p\) satisfies the full nonhomogeneous differential equation.

Initial Conditions

Solving differential equations often involves applying initial conditions to find a unique solution. Initial conditions are the values of the function and its derivatives at the start of the scenario being modeled, often at \( t = 0 \) or \( x = 0 \) for physical problems. They allow us to pinpoint the specific constants for the problem at hand.

With the given initial conditions \( y(0) = 0 \) and \( y'(0) = 1 \) for our exercise, we have the necessary information to determine the constants \( c1 \) and \( c2 \) for the general homogeneous solution. After finding the form of the general solution, we apply the initial conditions by substituting \( x = 0 \) and solving the resulting equations, thus defining the constants and refining our solution to one that reflects the particular situation described by those conditions.

General Solution of ODE

The general solution of an ODE (ordinary differential equation) is the combination of its homogeneous solution (related to the characteristic equation) and a particular solution (reflecting the nonhomogeneous part). It encapsulates all possible behaviors of the differential equation under various initial conditions and allows for the adaptation to specific scenarios once those conditions are defined.

To arrive at the general solution for our exercise, we add together the homogeneous solutions \( e^{x} \) and \( xe^{x} \) and the particular solution \( Ax^2e^{x} \). It is then represented as \( y(x) = c1*e^{x} + c2*xe^{x} + A*x^{2}*e^{x} \). This formula is designed to satisfy both the differential equation and the initial conditions provided, making it the bountiful reservoir that contains every specific solution that could emerge from the ODE, contingent on the initial conditions.