Question

Use variation of parameters to find a particular solution, given the solutions \(y_{1}, y_{2}\) of the complementary equation. $$ x^{2} y^{\prime \prime}-x(x+4) y^{\prime}+2(x+3) y=x^{4} e^{x} ; \quad y_{1}=x^{2}, \quad y_{2}=x^{2} e^{x} $$

Short answer

To summarize, we were given the differential equation \(x^2y'' - x(x+4)y' + 2(x+3)y = x^4e^x\) along with two complementary solutions, \(y_1 = x^2\) and \(y_2 = x^2e^x\). Using the method of variation of parameters, we found a particular solution to the differential equation: \(y_p = -\frac{1}{8}x^6 + \frac{1}{8}x^6e^x\).

Step-by-step solution

Write the given differential equation and complementary solutions in general form

We are given the following differential equation: $$ x^2y'' - x(x+4)y' + 2(x+3)y = x^4e^x $$ And the complementary solutions \(y_1 = x^2\) and \(y_2 = x^2e^x\).

Write down the Wronskian of \(y_{1}\) and \(y_{2}\)

Calculate the Wronskian, denoted as \(W(y_{1},y_{2})\), of the two complementary solutions. The Wronskian is given by the following determinant: $$ W(y_{1},y_{2}) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \\ \end{vmatrix} $$ Find the derivatives of \(y_1\) and \(y_2\): $$ y_1' = 2x, \quad y_2' = x^2e^x + 2xe^x $$ Now compute the Wronskian: $$ W(y_{1},y_{2}) = \begin{vmatrix} x^2 & x^2e^x \\ 2x & x^2e^x + 2xe^x \\ \end{vmatrix} = (x^2)(x^2e^x + 2xe^x) - (2x)(x^2e^x) = 2x^3e^x $$

Find the expressions for \(u_{1}'\) and \(u_{2}'\)

According to the method of variation of parameters, we can find expressions for \(u_{1}'\) and \(u_{2}'\) as follows: $$ u_1' = -\frac{y_2 f}{W(y_1, y_2)} \quad\text{and}\quad u_2' = \frac{y_1 f}{W(y_1, y_2)} $$ Where \(f(x) = x^4e^x\) is the function on the right side of the given differential equation. Substitute the given values of \(y_1\), \(y_2\), \(W(y_1, y_2)\), and \(f(x)\) into the equations: $$ u_1' = -\frac{x^2e^x(x^4e^x)}{2x^3e^x} \quad\text{and}\quad u_2' = \frac{x^2(x^4e^x)}{2x^3e^x} $$ Simplify the expressions: $$ u_1' = -\frac{1}{2}x^3 \quad\text{and}\quad u_2' = \frac{1}{2}x^3 $$

Integrate \(u_{1}'\) and \(u_{2}'\) to find \(u_{1}\) and \(u_{2}\)

Now, integrate \(u_1'\) and \(u_2'\) to find \(u_1(x)\) and \(u_2(x)\): $$ u_1(x) = \int u_1' dx = -\frac{1}{2} \int x^3 dx = -\frac{1}{8}x^4 + C_1 $$ $$ u_2(x) = \int u_2' dx = \frac{1}{2} \int x^3 dx = \frac{1}{8}x^4 + C_2 $$ Here, \(C_1\) and \(C_2\) are constants of integration.

Find a particular solution using \(y = u_{1}(x)y_{1} + u_{2}(x)y_{2}\)

Now that we have expressions for \(u_1(x)\) and \(u_2(x)\), we can find the particular solution by substituting these expressions into the following equation: $$ y_p = u_1(x)y_1 + u_2(x)y_2 $$ Substitute the expressions of \(u_1(x)\), \(u_2(x)\), \(y_1\), and \(y_2\): $$ y_p = \left(-\frac{1}{8}x^4 + C_1\right)x^2 + \left(\frac{1}{8}x^4 + C_2\right)x^2e^x $$ Since we only want a particular solution, we can ignore the constants of integration, so we get: $$ y_p = -\frac{1}{8}x^6 + \frac{1}{8}x^6e^x $$ So, the particular solution is: $$ y_p = -\frac{1}{8}x^6 + \frac{1}{8}x^6e^x $$

Key concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They play a crucial role in modeling real-world phenomena where change is continuous, such as in physics, engineering, and economics.

A differential equation can be classified by its order, defined by the highest derivative it contains. The given exercise involves a second-order differential equation because the highest derivative is the second derivative, denoted as \(y''\).

The form of the differential equation in the exercise, \(x^2y'' - x(x+4)y' + 2(x+3)y = x^4e^x\), demonstrates how differential equations can equate an unknown function \(y(x)\) and its derivatives to some function of \(x\). Solving this equation involves finding the function \(y(x)\) that satisfies the relationship.

Wronskian

The Wronskian is a determinant used in the theory of differential equations to determine whether two functions are linearly independent. This is important when finding the complementary solution to a homogeneous differential equation.

In practical terms, when the Wronskian of two solutions is non-zero on an interval, it implies they are linearly independent and form a fundamental set of solutions on that interval. In the provided step-by-step solution, the Wronskian for \(y_1 = x^2\) and \(y_2 = x^2e^x\) is computed as \(2x^3e^x\), which is non-zero for all \(x > 0\), confirming that \(y_1\) and \(y_2\) form a fundamental set of solutions for the homogeneous part of the differential equation.

Particular Solution

A particular solution to a differential equation is a solution that satisfies the non-homogeneous equation. It is specific to the form of the non-homogeneous term. In contrast to complementary or homogeneous solutions, which include arbitrary constants, a particular solution does not involve these constants and represents one instance of an infinite number of potential solutions that satisfy the equation.

In the context of the exercise, the method of variation of parameters is used to find a particular solution that takes into account the specific non-homogeneous term, \(x^4e^x\), thus providing a concrete function \(y_p(x)\) that fits into the original differential equation.

Complementary Solutions

Complementary solutions, also known as homogeneous solutions, are solutions to the homogeneous part of a differential equation, which is obtained by setting the non-homogeneous part to zero. The general solution to a linear differential equation is typically the sum of its complementary solution and a particular solution.

In the process outlined in the solution, \(y_1 = x^2\) and \(y_2 = x^2e^x\) represent the complementary solutions to the homogeneous differential equation. These solutions come with arbitrary constants, which are used to satisfy initial conditions or boundary values specific to a particular problem scenario.

Method of Undetermined Coefficients

The method of undetermined coefficients, also known as the method of educated guesses, is a technique to determine a particular solution to a non-homogeneous linear differential equation. This method assumes that the particular solution has a certain form, depending on the non-homogeneous term, and then it determines the unknown coefficients by substituting it into the differential equation.

However, when the non-homogeneous term is more complex or does not easily suggest a form for the particular solution, the method of variation of parameters, used in the provided exercise, is a more flexible alternative approach. Unlike the method of undetermined coefficients, variation of parameters does not require guessing the form of a particular solution and works for a broader class of non-homogeneity.