Question

Find a particular solution, given the fundamental set of solutions of the complementary equation. $$ 4 x^{3} y^{\prime \prime \prime}+4 x^{2} y^{\prime \prime}-5 x y^{\prime}+2 y=30 x^{2} ; \quad\left\\{\sqrt{x}, 1 / \sqrt{x}, x^{2}\right\\} $$

Short answer

Question: Find a particular solution to the given third-order linear differential equation, given a fundamental set of solutions for the complementary equation: Given: Fundamental set of solutions \(\left\{\sqrt{x}, 1 / \sqrt{x}, x^{2}\right\}\) and the equation is \(y'''-6y''+12y'-8y=30x^2\) Solution: From our calculations in the step by step solution, a particular solution for the given equation is: $$ y_p(x) = C_1(x) \sqrt{x} + C_2(x) \frac{1}{\sqrt{x}} + C_3(x) x^2 $$ where \(C_1(x)\), \(C_2(x)\), and \(C_3(x)\) are the values found after integrating the expressions found in Step 4.

Step-by-step solution

Write the general solution of the complementary equation

Using the given fundamental set of solutions \(\left\{\sqrt{x}, 1 / \sqrt{x}, x^{2}\right\}\), we can write the general solution of the complementary equation as: $$ y_c = C_1 \sqrt{x} + C_2 \frac{1}{\sqrt{x}} + C_3 x^2 $$

Set up the Wronskian

To set up the Wronskian, we will first find the derivatives of the given set of solutions, then compute the determinant using these derivatives. The derivatives are: $$ \begin{aligned} \frac{d}{dx} (\sqrt{x}) = \frac{1}{2\sqrt{x}}, \;\;\; \frac{d}{dx} \left(\frac{1}{\sqrt{x}}\right) = -\frac{1}{2x\sqrt{x}}, \;\;\; \frac{d}{dx} (x^2) = 2x \end{aligned} $$ Now compute the Wronskian, \(W\): $$ W = \begin{vmatrix} \sqrt{x} & 1/\sqrt{x} & x^2 \\ 1/(2\sqrt{x}) & -1/(2x\sqrt{x}) & 2x \\ (1-2x)/(4x\sqrt{x}) & 1/(4x^2\sqrt{x}) & 2 \end{vmatrix} $$

Write the variation of parameters formulas for the coefficients

Now, using the method of variation of parameters, we will write the coefficients \(C_1'(x), C_2'(x)\), and \(C_3'(x)\) as follows: $$ C_1'(x) = -\frac{A(x)}{W(x)}\frac{1}{\sqrt{x}}, \;\;\; C_2'(x) = \frac{A(x)}{W(x)}\sqrt{x}, \;\;\; C_3'(x) = -\frac{A(x)}{W(x)}\frac{1}{2x} $$ where \(A(x)\) is the term on the right-hand side of the given equation, which in this case is \(30x^2\).

Solve for the required coefficients using the given differential equation

Now we will solve for each individual parameter by integrating each equation as follows: $$ \begin{aligned} C_1(x) &= \int -\frac{30x^2}{W(x)}\frac{1}{\sqrt{x}} dx \\ C_2(x) &= \int \frac{30x^2}{W(x)}\sqrt{x} dx \\ C_3(x) &= \int -\frac{30x^2}{W(x)}\frac{1}{2x} dx \end{aligned} $$ After finding the values for \(C_1(x)\), \(C_2(x)\), and \(C_3(x)\), we can rewrite the general solution of the complementary equation.

Find the particular solution

Now, we can insert the computed \(C_1(x)\), \(C_2(x)\), and \(C_3(x)\) into the complementary equation to obtain the particular solution: $$ y_p(x) = C_1(x) \sqrt{x} + C_2(x) \frac{1}{\sqrt{x}} + C_3(x) x^2 $$ Finally, the particular solution for the given equation is found by substituting the values of the coefficients found in the previous step.

Key concepts

Fundamental Set of Solutions

In solving differential equations, a fundamental set of solutions plays a crucial role. This set consists of solutions that form the building blocks for the general solution of a differential equation. For a linear differential equation with constant coefficients, the fundamental set includes as many functions as the order of the differential equation. This occurs because these solutions are linearly independent. For example, in our exercise, the fundamental set of solutions draws from the given functions:

• \( \sqrt{x} \)
• \( \frac{1}{\sqrt{x}} \)
• \( x^2 \)

Each solution in the set satisfies the complementary equation, which is the homogeneous version of the differential equation we aim to solve. By combining these solutions linearly, one can construct the general solution of the complementary differential equation:
\[ y_c = C_1 \sqrt{x} + C_2 \frac{1}{\sqrt{x}} + C_3 x^2 \]
Where \( C_1, C_2, \) and \( C_3 \) are arbitrary constants determined by initial conditions or boundary values.

Variation of Parameters

Variation of parameters is a method used to find particular solutions to non-homogeneous differential equations. This technique complements the solution obtained from the homogeneous equation. While it may seem a bit complex at first, its core idea lies in allowing the constants of the complementary solution to become functions. This adjustment lets these 'parameters' vary with the independent variable, usually \( x \).
Here's how we generally set it up using our exercise:

• The derivatives of these coefficients transform into expressions that incorporate the original non-homogeneous term, \( A(x) \).
• These expressions, \( C_1'(x), C_2'(x), \) and \( C_3'(x), \) are expressed in terms of the Wronskian and original equation coefficients.

The goal here is to integrate these expressions to find \( C_1(x), C_2(x), \) and \( C_3(x), \) which plug back into the fundamental solutions:

• \( C_1(x) \sqrt{x} \)
• \( C_2(x) \frac{1}{\sqrt{x}} \)
• \( C_3(x) x^2 \)

This method provides the particular solution, adding it to the solution of the homogeneous equation offers the complete solution of the differential equation.

Wronskian Determinant

The Wronskian is a determinant used to test the linear independence of a set of solutions to a differential equation. A non-zero Wronskian at a certain point suggests that the functions in your set are linearly independent, making them a valid fundamental set.
Calculating the Wronskian involves differentiating the given solutions, arranging them into a matrix, and then calculating the determinant. For our example:

• We begin with solutions \( \sqrt{x}, \frac{1}{\sqrt{x}}, x^2 \) and their first derivatives.
• These are placed into a matrix to set up the Wronskian.
• The determinant of this matrix gives us the Wronskian, \( W \).

In the exercise shown, computing the Wronskian forms part of the variation of parameters method because it appears in denominators when solving for the particular solution. It ensures that the solution set spans the solution space which is crucial for accurately solving the differential equation.

Complementary Equation

The complementary equation is the homogeneous version of the differential equation. This means it equals zero and excludes the non-homogeneous part on the right-hand side. Solving the complementary equation is often the first step in tackling a non-homogeneous differential equation.
To solve, you need a fundamental set of solutions for this homogeneous equation. For the given exercise, having the complementary equation with the fundamental set

• \( \sqrt{x} \)
• \( \frac{1}{\sqrt{x}} \)
• \( x^2 \)

enables us to form its general solution:
\[ y_c = C_1 \sqrt{x} + C_2 \frac{1}{\sqrt{x}} + C_3 x^2 \]
These solutions help construct the overall solution, giving context before integrating with the particular solution derived through methods such as the variation of parameters. This approach is essential for understanding how both solutions come together to solve non-homogeneous differential equations comprehensively.