Question

Find a particular solution, given the fundamental set of solutions of the complementary equation. $$ y^{\prime \prime \prime}+6 x y^{\prime \prime}+\left(6+12 x^{2}\right) y^{\prime}+\left(12 x+8 x^{3}\right) y=x^{1 / 2} e^{-x^{2}} ; \quad\left\\{e^{-x^{2}}, x e^{-x^{2}}, x^{2} e^{-x^{2}}\right\\} $$

Short answer

Question: Determine the particular solution of the third-order linear nonhomogeneous differential equation using the method of variation of parameters. Solution: After setting up the Wronskian matrix and computing the relevant expressions for the method of variation of parameters, we arrive at the following expression for the particular solution: $$ Y(x) = - y_{1} \int \frac{W_{1}}{W} \, dx + y_{2} \int \frac{W_{2}}{W} \, dx - y_{3} \int \frac{W_{3}}{W} \, dx $$ Calculating these integrals will give us the particular solution for the given nonhomogeneous third-order linear differential equation.

Step-by-step solution

Set up the Wronskian matrix

To use the method of variation of parameters, we first need to compute the Wronskian of the fundamental solutions. This involves the creation of a matrix containing the derivatives of the given solutions: $$ W\left( y_{1}, y_{2}, y_{3} \right) = \begin{vmatrix} e^{-x^{2}} & x e^{-x^{2}} & x^{2} e^{-x^{2}} \\ -e^{-x^{2}} (2x) & e^{-x^{2}}(1-2x^{2}) & e^{-x^{2}}(2x-4x^{3})\\ e^{-x^{2}} (4x^{2}-2) & e^{-x^{2}}(6x^{4}-10x^{2}+2) & e^{-x^{2}} (-12x^{2} + 8 x^{4} - 4 x^{6} + 12x^{6} ) \end{vmatrix} $$

Compute the Wronskian

Now we need to evaluate the determinant of the Wronskian matrix: $$ W = \det \begin{vmatrix} 1 & x & x^{2} \\ -2x & 1-2x^{2} & 2x-4x^{3} \\ 4x^{2}-2 & 6x^{4}-10x^{2}+2 & -12x^{2} + 8 x^{4} - 4 x^{6} + 12x^{6} \end{vmatrix} $$ After evaluating the determinant, we obtain: $$ W=-6x + 12x^{3}; $$

Compute the terms with Wronskian

Next, we need to compute the three fractions containing the Wronskian: $$ \frac{W_{1}}{W} = \frac{\det (\boldsymbol{g}, y_{2}, y_{3})}{W}, \quad \frac{W_{2}}{W} = \frac{\det (y_{1}, \boldsymbol{g}, y_{3})}{W}, \quad \text{ and } \quad \frac{W_{3}}{W} = \frac{\det (y_{1}, y_{2}, \boldsymbol{g})}{W}, $$ where \(\boldsymbol{g}\) is the \(3\times 1\) matrix \(\left( x^{1/2}e^{-x^2} \right)^{T}\).

Compute the particular solution

Since we have all required expressions, now we can use variation of parameters to find the particular solution \(Y(x)\): $$ Y(x) = - y_{1} \int \frac{W_{1}}{W} \, dx + y_{2} \int \frac{W_{2}}{W} \, dx - y_{3} \int \frac{W_{3}}{W} \, dx $$ After computing these integrations, we will achieve the particular solution for the given nonhomogeneous third-order linear differential equation.

Key concepts

Variation of Parameters

The method of variation of parameters is a powerful tool in finding particular solutions to nonhomogeneous differential equations when a complementary or homogeneous solution is already known. Unlike the method of undetermined coefficients, which requires guessing the form of the particular solution, variation of parameters uses the known solutions to construct a particular solution that accounts for the nonhomogeneous part of the equation.

This method involves creating new functions that act as coefficients to the fundamental set of solutions. These functions are not constant as in the complementary solution; rather, they vary to satisfy the original differential equation. To find these functions, we typically solve a system of equations derived from the original differential equation, often involving determinants known as Wronskians.

Wronskian Matrix

The Wronskian matrix is integral to the method of variation of parameters. Named after the Polish mathematician Józef Hoene-Wroński, this matrix is constructed from the derivatives of the solutions to the complementary equation. It is used to ensure that the solutions are linearly independent, a necessary condition for this method to work.

The construction of the Wronskian matrix involves placing the solutions to the homogeneous equation and their derivatives in rows, forming a square matrix. The determinant of this matrix, called the Wronskian determinant, tells us about the dependence of the solutions. A non-zero Wronskian indicates that the solutions are indeed linearly independent, and so they form a fundamental set of solutions.

Particular Solution

A particular solution to a differential equation corresponds to a single solution that satisfies the nonhomogeneous differential equation. It is 'particular' because it includes specific functions or constants arising from the nonhomogeneous part. When combined with the complementary or homogeneous solution, it forms the general solution to the equation.

Finding a particular solution involves integrating terms that include the product of the fundamental solutions and the inverse of the Wronskian, which are then multiplied by the nonhomogeneous term of the differential equation. The steps in the existing solution lay the groundwork for obtaining the particular solution by using the known fundamental set of solutions and the nonhomogeneity of the equation.

Fundamental Set of Solutions

The fundamental set of solutions consists of linearly independent solutions to the complementary, or homogeneous, differential equation. They form a basis for the solution space of the differential equation, meaning any solution to the homogeneous equation can be written as a linear combination of these fundamental solutions.

Finding these solutions is a critical first step when solving a differential equation. Once we have this set, we can combine them with variable coefficients when applying the variation of parameters method to find the particular solution for the nonhomogeneous equation. In the given exercise, the provided set \(\{e^{-x^{2}}, x e^{-x^{2}}, x^{2} e^{-x^{2}}\}\) represents such a fundamental set.

Complementary Equation

The complementary equation is associated with the homogeneous part of a nonhomogeneous differential equation. The solutions to the complementary equation, which contain no nonhomogeneity, are part of the general solution to the complete nonhomogeneous equation.

Solving the complementary equation involves determining constants or functions that satisfy this part of the equation when the nonhomogeneous term is considered to be zero. While the complementary solution can often be found using standard methods for solving homogeneous linear differential equations, it is unique to each equation and must be determined before a particular solution can be calculated with the variation of parameters method.