Question

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

Short answer

Question: Determine the particular solution for the given inhomogeneous differential equation with the given fundamental set of solutions for the related complementary equation: Differential equation: \(x^2 y''' + 2x y'' - (x^2 + 2) y' = 2x^2\) Fundamental set of complementary solutions: \(\left\{1, e^{x} / x, e^{-x} / x\right\}\) Answer: The particular solution for the given inhomogeneous differential equation is \(y_p(x) = x^2 - 1\).

Step-by-step solution

Identifying the Differential Equation Type and Guessing The Form of Particular Solution

The given equation is a third-order linear weight-homogeneous differential equation, as all the coefficients are weights of the unknowns (derivations of y). We know that the fundamental set of solutions for the related complementary equation is \(\left\\{1, e^{x} / x, e^{-x} / x\right\\}\), and the right-hand side of the given equation is of the form \(2x^2\). Since, the right-hand side of the equation does not match any functions in the complementary solutions set, we can directly guess a polynomial of same degree for our particular solution. Let's assume, $$ y_p(x) = Ax^2 + Bx + C $$ Next, we need to find the derivatives of the proposed particular solution.

Finding the Derivatives of the Proposed Particular Solution

In order to find the undetermined coefficients of the assumed particular solution, we need to substitute it into the differential equation. So, we have to find the first, second, and third derivatives of \(y_p(x)\). First derivative: $$ y_p'(x) = 2Ax + B $$ Second derivative: $$ y_p''(x) = 2A $$ Third derivative: $$ y_p'''(x) = 0 $$

Substituting the Particular Solution into the Differential Equation

Now we'll plug the particular solution (\(y_p\)) and its derivatives into the given differential equation to find the undetermined coefficients. We have: $$ x^2 (0) + 2x(2A) - (x^2 + 2)(2Ax + B) = 2x^2 $$ Simplifying the equation, we get: $$ 4Ax^2 - 2A x^2 - 2Bx - 4Ax - 2B = 2x^2 $$ Now, group the terms by x's power: $$ (4A - 2A)x^2 + (-2B - 4A)x - 2B = 2x^2 $$

Solving for the Undetermined Coefficients

Now, comparing the coefficients of the terms with the same power of x, we can set up a system of equations to solve for A, B, and C. $$ \begin{cases} 4A - 2A = 2 \Rightarrow 2A = 2 \\ -2B - 4A = 0 \\ -2B = 0 \end{cases} $$ Solving the system: $$ A = 1 \\ B = 0 \\ C = -1 $$ Thus, the particular solution is: $$ y_p(x) = x^2 - 1 $$

Construct the General Solution

Finally, we'll form the general solution by combining the particular solution with the fundamental set of solutions. The general solution for this non-homogeneous differential equation is: $$ y(x) = C_1 + C_2 \frac{e^x}{x} + C_3 \frac{e^{-x}}{x} + x^2 - 1 $$ Where \(C_1\), \(C_2\), and \(C_3\) are arbitrary constants.

Key concepts

Third-Order Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. When it comes to third-order differential equations, such equations involve the unknown function y, its derivatives y', y'', and the third derivative y'''.

Examples of physical systems where these equations may arise include certain motion scenarios under the influence of forces or torques, or electrical circuits with inductors, resistors, and capacitors interacting over time. To solve a third-order differential equation, it's crucial to recognize the type of differential equation we are dealing with because the solution approach can vary significantly.

The exercise provided showcases a third-order linear weight-homogeneous differential equation. Its standard form is a(x)y''' + b(x)y'' + c(x)y' + d(x)y = f(x), where f(x) is the source term or 'forcing function'. This type of equation admits a solution that can be expressed as a sum of a particular solution (y_p) of the non-homogeneous equation and a general solution (y_c) to the related homogeneous equation.

Particular Solution of Differential Equation

In any linear differential equation, finding a particular solution is about identifying a solitary specific solution that satisfies the equation. Unlike the general solution, which represents a family of possible solutions characterized by arbitrary constants, the particular solution addresses the non-homogeneous part of the equation directly.

To find such a solution, we often make an educated guess called the 'method of undetermined coefficients' or 'ansatz'. This entails guessing a form for the particular solution that has 'enough room' to fit the non-homogeneous part of the equation, and then determining any unknown coefficients.

In our step-by-step solution, we made a guess that a polynomial of the same degree as the right-hand side (the source term of 2x^2) would serve as a suitable particular solution. After finding the derivatives of our guess and plugging them into the original equation, we're able to solve for the coefficients A, B, and C. This process aligns perfectly with the exercise improvement advice of focusing on understanding the method used to find the particular solution.

Complementary Equation Solutions

When we refer to complementary equation solutions, or the homogeneous solutions, we're discussing a set of solutions that comprises the general solution (y_c) of the complementary or homogeneous version of the differential equation (where the right-hand side is zero). A set of functions that form a basis for the solution space of this homogeneous equation is called the fundamental set of solutions.

For our exercise, the fundamental set provided is \(\{1, e^{x} / x, e^{-x} / x\}\), which are the solutions to the homogeneous equation corresponding to our non-homogeneous differential equation.

The idea is to express the general solution to the non-homogeneous equation as a linear combination of these complementary solutions plus the particular solution. Thus, a proper understanding of homogeneous solutions forms the foundation for handling more complex differential equations where the non-homogeneous part can complicate things. Our final step combines the complementary solutions with the particular solution we found earlier, following the superposition principle, which is vital to solving linear homogeneous and non-homogeneous equations alike.