Question

Use variation of parameters to find a particular solution, given the solutions \(y_{1}, y_{2}\) of the complementary equation. $$ x^{2} y^{\prime \prime}-2 x(x+1) y^{\prime}+\left(x^{2}+2 x+2\right) y=x^{3} e^{x} ; \quad y_{1}=x e^{x}, \quad y_{2}=x^{2} e^{x} $$

Short answer

Based on the given information and the previous calculations, the particular solution of the differential equation can be expressed as: $$ y = u(x) (xe^x) + v(x) (x^2 e^x) $$ Where: $$ u(x) = -\int u(x) dx + \int c_1 dx $$ $$ v(x) = -\dfrac{c_2}{2} \ln|x| - \dfrac{c_2}{x} + \dfrac{c_2}{2}\ln|2+x| + c_3 $$ To find the actual function for the particular solution, we would need more information about initial conditions or boundary conditions. However, this general expression represents the form of the particular solution for the given second-order nonhomogeneous differential equation using the variation of parameters method.

Step-by-step solution

Write the particular solution using the given complementary solutions

We can express the particular solution as: $$ y = u(x) y_1 + v(x) y_2 = u(x)(xe^x) + v(x)(x^2 e^x) $$

Find the first and second derivatives of \(y\)

The first and second derivatives of \(y\) can be computed as follows: $$ y' = u'(x) y_1 + u(x) y_1' + v'(x) y_2 + v(x) y_2' $$ Since \(y_1' = (1+x)e^x\) and \(y_2' = e^x(2x+x^2)\), we have: $$ y' = u'(x)(xe^x) + u(x)(1+x)e^x + v'(x)(x^2 e^x) + v(x)e^x(2x+x^2) $$ And the second derivative: $$ y'' = u''(x) y_1 + 2u'(x) y_1' + u(x) y_1'' + v''(x) y_2 + 2v'(x) y_2' + v(x) y_2'' $$ Since \(y_1'' = (2+2x)e^x\) and \(y_2'' = 3xe^x(2+x)\), we get: $$ y'' = u''(x)(xe^x) + 2u'(x)(1+x)e^x + u(x)(2+2x)e^x + v''(x)(x^2 e^x) + 2v'(x)e^x(2x+x^2) + v(x)3xe^x(2+x) $$

Substitute \(y\), \(y'\), and \(y''\) into the given differential equation

Placing \(y\), \(y'\), and \(y''\) in the given differential equation results in: $$ x^2 (u''(x)(xe^x) + 2u'(x)(1+x)e^x + u(x)(2+2x)e^x + v''(x)(x^2 e^x) + 2v'(x)e^x(2x+x^2) + v(x)3xe^x(2+x)) \\ - 2x (x+1)(u'(x)(xe^x) + u(x)(1+x)e^x + v'(x)(x^2 e^x) + v(x)e^x(2x+x^2)) \\ + (x^2+2x+2)(u(x)(xe^x) + v(x)(x^2 e^x)) = x^3 e^x $$

Simplify the equation and solve for \(u'\) and \(v'\)

By equating the coefficients of the exponential function e^x and cancel the common terms, we obtain the following system of linear equations: $$ \begin{cases} xu''(x) + 2u'(x)(1+x) + (2+2x)u(x) + 3x(2+x)v(x) = x^2 \\ x^2v''(x) + 2x(2+x)v'(x) = 0 \end{cases} $$ Now, to solve this system, we integrate the first equation with respect to \(x\) to find \(u'(x)\), and the second equation to find \(v'(x)\). This results in: $$ \begin{cases} u'(x) = -u(x) + c_1 \\ v'(x) = \dfrac{c_2}{x^2(2+x)} \end{cases} $$

Integrate \(u'(x)\) and \(v'(x)\) to obtain \(u(x)\), and \(v(x)\)

The integral for \(u'(x)\) is straightforward, $$ u(x) = -\int u(x) dx + \int c_1 dx $$ For \(v'(x)\), use partial fraction decomposition, $$ v'(x) = \dfrac{c_2}{x^2(2+x)} = \dfrac{A}{x} + \dfrac{B}{x^2} + \dfrac{C}{2+x} $$ Solving for A, B, and C, we obtain: $$ A = -\dfrac{c_2}{2},\ B = c_2,\ C = \dfrac{c_2}{2} $$ Now, we can integrate \(v'(x)\): $$ v(x) = \int v'(x) dx = -\dfrac{c_2}{2} \int \dfrac{1}{x} dx + c_2 \int \dfrac{1}{x^2} dx + \dfrac{c_2}{2} \int \dfrac{1}{2+x} dx $$ This results in: $$ v(x) = -\dfrac{c_2}{2} \ln|x| - \dfrac{c_2}{x} + \dfrac{c_2}{2}\ln|2+x| + c_3 $$ By solving the integrals and replacing \(u(x)\) and \(v(x)\) into the particular solution, we will find the particular solution of the given differential equation.

Key concepts

Differential Equations

Differential equations are mathematical equations that describe how a function varies with respect to a variable. They involve derivatives, which represent rates of change. In essence, they are used to model real-world phenomena, involving quantities that change continuously. For example, processes like population growth, heat distribution, and mechanical vibrations can all be modeled using differential equations.

In solving these equations, we often aim to find a function, or a set of functions, that satisfies the given equation. There are various methods for solving differential equations, ranging from analytical to numerical approaches. Each method is chosen based on the type and complexity of the equation at hand.

Complementary Solution

A complementary solution of a differential equation is a solution that accounts for the homogeneous part of the equation. When the right-hand side of the differential equation equals zero, the resulting homogenous equation is solved to find the complementary solution.

The complementary solution, often denoted as the sum of functions like \(y_{1}(x)\) and \(y_{2}(x)\), characterizes the system's inherent behavior without external influence. In our exercise, the complementary functions are \(y_{1}=xe^x\) and \(y_{2}=x^2e^x\). These functions serve as a fundamental solution set for the differential equation when external forces (like our \(x^3e^x\) term) are absent.

Particular Solution

In contrast to the complementary solution, a particular solution addresses the non-homogeneous part of the differential equation. This includes the external forces or influences, represented by non-zero terms on the right-hand side of the equation.

The particular solution is crucial because it describes the response of the system specifically due to those external elements. In variation of parameters, we express the particular solution as a linear combination of the complementary solutions \(y = u(x) y_1 + v(x) y_2\). Our goal is to determine the functions \(u(x)\) and \(v(x)\) such that when they are substituted back into the original differential equation, the equation holds. This approach is particularly useful when the non-homogeneous term is more complex, as in our exercise with the term \(x^3e^x\).

Integration Techniques

Integration is a fundamental tool in finding particular solutions of differential equations, especially using the variation of parameters.

In this method, we find functions \(u'(x)\) and \(v'(x)\) that make our equation hold true. These functions are then integrated to find \(u(x)\) and \(v(x)\).

• For \(u'(x)\), direct integration might be straightforward. When given as \(-u(x) + c_1\) in the solution, an integration of both sides helps determine \(u(x)\).
• For \(v'(x)\), partial fraction decomposition can be employed. Decomposing \(v'(x)\) into simpler fractions facilitates easier integration. Solving these integrals provides us with the function \(v(x)\).

Particularly with complicated terms like \(x^2(2+x)\) in the denominator, breaking them into simpler components is essential. The solutions of these integrations are then plugged back into our expression for the particular solution to arrive at a complete solution of the differential equation.