Question

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

Short answer

Question: Determine the particular solution for the given second-order linear homogeneous differential equation using the method of variation of parameters: \(y'' + \frac{4}{x}y' + \frac{4}{x^2}y = x^{\frac{3}{2}}\), with complementary solutions \(y_1 = \frac{1}{x}\) and \(y_2 = x^3\). Answer: The particular solution for the given differential equation is: $$ y_p = -\left(\frac{1}{18}\ln|(3x^6 + x^{-6})|\right)x^{\frac{3}{2}} + \left(\frac{1}{6}\ln|(3x^6 + x^{-6})|\right)x^6 $$

Step-by-step solution

Write down the complementary solutions

We are given the complementary solutions \(y_1 = \frac{1}{x}\) and \(y_2 = x^3\). We'll use these solutions to set up the variation of parameters method.

Find the Wronskian of the complementary solutions

We need to find the Wronskian, \(W(y_1, y_2)\), which is defined as: $$ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} $$ Compute the derivatives of \(y_1\) and \(y_2\): $$ y_1' = -\frac{1}{x^2}, \quad y_2' = 3x^2 $$ Now compute the Wronskian: $$ W(y_1, y_2) = \begin{vmatrix} \frac{1}{x} & x^3 \\ -\frac{1}{x^2} & 3x^2 \end{vmatrix} = 3x^3 + \frac{1}{x^3} $$

Compute the particular solution

Using the variation of parameters method, we can write the particular solution \(y_p\) as: $$ y_p = -y_1 \int \frac{y_2 Q(x)}{W(y_1, y_2)} dx + y_2 \int \frac{y_1 Q(x)}{W(y_1, y_2)} dx $$ where \(Q(x)\) is the right-hand side of the nonhomogeneous equation, which in our case is \(x^{\frac{3}{2}}\). Now we can substitute the given functions and the Wronskian into the formula: $$ y_p = -\frac{1}{x} \int \frac{x^3 x^{\frac{3}{2}}}{3x^3 + \frac{1}{x^3}} dx + x^3 \int \frac{\frac{1}{x} x^{\frac{3}{2}}}{3x^3 + \frac{1}{x^3}} dx $$ Simplify the integrals: $$ y_p = -\frac{1}{x} \int \frac{x^{\frac{9}{2}}}{3x^6 + x^{-6}} dx + x^3 \int \frac{x^{\frac{1}{2}}}{3x^6 + x^{-6}} dx $$

Evaluate the integrals

Now we need to evaluate the integrals. We notice that the integrands have the same common denominator, which will simplify our calculations. $$ y_p = -\frac{1}{x} \int \frac{x^{\frac{9}{2}}}{3x^6 + x^{-6}} dx + x^3 \int \frac{x^{\frac{1}{2}}}{3x^6 + x^{-6}} dx $$ Combine the integrals into one: $$ y_p = \int \left[-\frac{1}{x} \frac{x^{\frac{9}{2}}}{3x^6 + x^{-6}} + x^3 \frac{x^{\frac{1}{2}}}{3x^6 + x^{-6}}\right] dx $$ Now, substitute \(u = 3x^6 + x^{-6}\), hence \(du = (18x^5 - 6x^{-7})dx\). We can rewrite the integral in terms of \(u\) as: $$ y_p = \int \frac{-x^{\frac{3}{2}} + x^6}{u} du $$ We can now evaluate this integral by splitting the fraction into two parts: $$ y_p = -\int \frac{x^{\frac{3}{2}}}{u} du + \int \frac{x^6}{u} du $$ Perform a substitution to solve each integral: $$ y_p = -\left(\frac{1}{18}\ln|u|\right)x^{\frac{3}{2}} + \left(\frac{1}{6}\ln|u|\right)x^6 $$ Now substitute back for \(u\) to obtain: $$ y_p = -\left(\frac{1}{18}\ln|(3x^6 + x^{-6})|\right)x^{\frac{3}{2}} + \left(\frac{1}{6}\ln|(3x^6 + x^{-6})|\right)x^6 $$ This is the particular solution for the given differential equation.

Key concepts

Particular Solution

In the context of differential equations, a particular solution is a solution to a nonhomogeneous differential equation that accounts for the nonhomogeneity, or the presence of a non-zero right-hand side of the equation. When solving differential equations, it is common to divide the process into finding the complementary solution and the particular solution.
The particular solution specifically addresses the nonhomogeneous component of the equation. That component often arises from external forces and inputs. For a typical nonhomogeneous differential equation expressed as\[ ax^2 y'' + bxy' + cy = f(x) \]we aim to find a function, known as the particular solution, which satisfies this equation. It does so when added to the complementary solution. These combined solutions form the general solution to the equation.
In this particular case, using the method of variation of parameters, we expressed the particular solution as a combination of integrals. We utilized the complementary solutions (\(y_1 = \frac{1}{x}\) and \(y_2 = x^3\)) to construct and solve these integrals.

Complementary Solution

Every nonhomogeneous differential equation has an associated homogeneous equation. The complementary solution is the solution to this homogeneous equation. It is an essential part of the general solution.
In simpler terms, the complementary solution solves the differential equation without considering the additional nonhomogeneous part (external forces). For the given differential equation:\[ x^{2} y'' - x y' - 3y = 0 \]the complementary solution can be represented as a combination of linearly independent solutions. Here, these solutions are given as:

• \(y_1 = \frac{1}{x}\)
• \(y_2 = x^3\)

A linear combination of these solutions (say, \(c_1y_1 + c_2y_2\) where \(c_1\) and \(c_2\) are constants) will form the complementary solution which is essential for the general solution. These solutions describe the behavior of the system in the absence of external forces.

Wronskian

The Wronskian is a special determinant used in solving differential equations. It helps assess whether a set of functions (solutions to a differential equation) are linearly independent. This property of independence is crucial in forming a complete solution set.
For our given complementary solutions, the Wronskian is calculated using their derivatives:\[ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} \]Substituting the derivatives of \(y_1 = \frac{1}{x}\) and \(y_2 = x^3\), we get:\[ y_1' = -\frac{1}{x^2}, \quad y_2' = 3x^2 \]So the Wronskian becomes:\[ W(y_1, y_2) = \begin{vmatrix} \frac{1}{x} & x^3 \ -\frac{1}{x^2} & 3x^2 \end{vmatrix} = 3x^3 + \frac{1}{x^3} \]
The non-zero result of the Wronskian confirms that \(y_1\) and \(y_2\) are indeed linearly independent.

Nonhomogeneous Differential Equation

A nonhomogeneous differential equation is one where the equation is not equal to zero. It includes an additional component or function on its right-hand side signifying external forces acting upon the system.
This is notably different from a homogeneous differential equation where everything equates to zero. The general framework for our nonhomogeneous differential equation is:\[ x^{2} y'' - x y' - 3y = x^{3/2} \]
In this equation, the term \(x^{3/2}\) represents the nonhomogeneity and acts as the external input you're solving the equation in relation to. Solving such equations involves finding both the complementary and particular solutions. The process focuses on using these components to develop a general solution that fits the entire equation, including the nonhomogeneous part.