Question

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

Short answer

Question: Find the particular solution of the given third-order nonhomogeneous linear ordinary differential equation when the fundamental set of solutions of the corresponding complementary equation is given as \(\{x, x^{2}, 1/x\}\): $$ x^3y''' + x^2y'' - 2xy' + 2y = x^2 $$ Answer: The particular solution for the given nonhomogeneous linear ordinary differential equation is: $$ y_p(x) = \frac{1}{2}x^2 $$

Step-by-step solution

Identify the form of the particular solution and introduce the undetermined coefficients

Since the nonhomogeneous term has the form of a polynomial of degree 2, we will guess that the particular solution has the same form. Let the particular solution be: $$ y_p(x) = Ax^2 + Bx + C $$ where A, B, and C are the undetermined coefficients.

Calculate the derivatives of the guessed solution

In order to insert the guess into the differential equation, we need to calculate the first, second, and third derivatives of the guessed function. For \(y_p(x) = Ax^2 + Bx + C\), the derivatives are: $$ y_p'(x) = 2Ax + B \\ y_p''(x) = 2A \\ y_p'''(x) = 0 $$

Substitute the guessed solution and its derivatives into the differential equation

Now we substitute our guessed solution and its derivatives into the given differential equation. $$ x^3(0) + x^2(2A) - 2x(2Ax + B) + 2(Ax^2 + Bx + C) = x^2 $$ Simplify the equation: $$ 2Ax^3 - 4Ax^2 - 2Bx + 2Ax^2+ 2Bx + 2C = x^2 $$

Match the coefficients in the equation

For the equation to hold true for all x values, the coefficients on both sides of the equation must match. By comparing the coefficients, we obtain: $$ 2A = 1 \\ -4A + 2A = 0 \\ 2C = 0 $$

Solve for the undetermined coefficients

Now, we solve for the undetermined coefficients A, B, and C. $$ 2A = 1 \quad\Rightarrow\quad A = \frac{1}{2} \\ -4A + 2A = 0 \quad\Rightarrow\quad -2A = 0 \quad\Rightarrow\quad A = 0 \\ 2C = 0 \quad\Rightarrow\quad C = 0 $$

Write down the particular solution

With the determined coefficients, we now write down the particular solution for the given nonhomogeneous differential equation: $$ y_p(x) = \frac{1}{2}x^2 + 0x + 0 = \frac{1}{2}x^2 $$ Now we have the particular solution. To obtain the general solution of the nonhomogeneous equation, we need to combine the homogeneous solutions with this particular solution. The general solution would have the form: $$ y(x) = c_1x + c_2x^2 + c_3\frac{1}{x} + \frac{1}{2}x^2 $$ where \(c_1\), \(c_2\), and \(c_3\) are constants determined by the initial conditions.

Key concepts

Nonhomogeneous Differential Equation

A nonhomogeneous differential equation includes a term that does not depend on the function being solved, usually referred to as the 'nonhomogeneous term'. In the equation\[x^{3} y^{\prime \prime \prime}+x^{2} y^{\prime \prime}-2 x y^{\prime}+2 y = x^{2},\]the right side, \(x^2\), makes it nonhomogeneous.
Nonhomogeneous differential equations can be challenging because the solution must satisfy both the homogeneous and nonhomogeneous parts. Here, a combination of general solutions to the associated homogeneous equation and a particular solution to the nonhomogeneous equation helps solve the problem. The general solution of the differential equation is a sum of these two parts.

Particular Solution

The particular solution is a specific solution that satisfies the nonhomogeneous differential equation, particularly the nonhomogeneous term. For this exercise, the goal was to find a function \(y_p(x)\) such that when substituted into the differential equation, both sides of the equation are equal. The form \(y_p(x) = Ax^2 + Bx + C\) takes on a polynomial structure, matching the degree of the nonhomogeneous term.
Before finding the solution, we calculate derivatives of the guessed function and substitute them back into the equation.
This involves breaking down the equation into terms, comparing coefficients, and solving for the coefficients \(A\), \(B\), and \(C\). Successfully finding these coefficients allows us to write down the particular solution for the differential equation.

Undetermined Coefficients

The method of undetermined coefficients is used to find particular solutions for linear nonhomogeneous differential equations when the nonhomogeneous term is a polynomial, exponential, sine, or cosine function. This approach involves:

• Guessing a form for the particular solution based on the nonhomogeneous term.
• Determining the coefficients by substituting the guessed solution back into the differential equation.

In our exercise, the nonhomogeneous term \(x^2\) suggests a quadratic form for \(y_p(x)\). After guessing \(y_p(x) = Ax^2 + Bx + C\), we derive and substitute these expressions into the equation to solve for \(A\), \(B\), and \(C\).
This systematic approach reduces complex problems into more manageable algebraic steps.

Fundamental Set of Solutions

The fundamental set of solutions refers to the basic solutions of the associated homogeneous differential equation, which in our problem is the equation without the \(x^2\). Given the set \(\{x, x^2, 1/x\}\), these solutions form the complementary solution. They can be combined linearly to solve the homogeneous equation:
\[ y(x) = c_1x + c_2x^2 + c_3\frac{1}{x}, \]where \(c_1\), \(c_2\), and \(c_3\) are arbitrary constants. These constants are determined by initial conditions or boundary values.
By combining this fundamental set with the particular solution, we can express the general solution to the original nonhomogeneous equation, allowing for a complete understanding of all potential solutions.