Problem 84 - Calculus 3

Question

In each of the following problems, two linearly independent solutions - \(y_{1}\) and \(y_{2}-\) are given that satisfy the corresponding homogeneous equation. Use the method of variation of parameters to find a particular solution to the given nonhomogeneous equation. Assume \(x > 0\) in each exercise. \(x^{2} y^{\prime \prime}+2 x y^{\prime}-2 y=3 x\), \(y_{1}(x)=x, y_{2}(x)=x^{-2}\)

Step-by-Step Solution

Answer

The particular solution is \( y_p(x) = \frac{x^5 - x^2}{4} \).

Step 1: Write Auxiliary Functions

Given that the homogeneous solutions are \( y_1(x) = x \) and \( y_2(x) = x^{-2} \), they are both solutions of the homogeneous equation. According to the variation of parameters method, we first need to define two auxiliary functions, \( u_1(x) \) and \( u_2(x) \), which we'll use as follows:\[y_p(x) = u_1(x)y_1(x) + u_2(x)y_2(x)\]

Step 2: Use Wronskian to Find Formulas for \(u_1'\) and \(u_2'\)

The Wronskian \( W \) of \( y_1 \) and \( y_2 \) is given by:\[W = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} = \begin{vmatrix} x & x^{-2} \ 1 & -2x^{-3} \end{vmatrix} = -3x^{-2}\]Using the Wronskian, the derivatives of \( u_1(x) \) and \( u_2(x) \) can be formulated:\[u_1' = \frac{-y_2 f(x)}{W} = \frac{-x^{-2} \cdot 3x}{-3x^{-2}} = x^3\]\[u_2' = \frac{y_1 f(x)}{W} = \frac{x \cdot 3x}{-3x^{-2}} = -x^3\]

Step 3: Integrate to Find \(u_1\) and \(u_2\)

Integrate the expressions for \( u_1'(x) \) and \( u_2'(x) \) to find \( u_1(x) \) and \( u_2(x) \):\[u_1(x) = \int x^3 dx = \frac{x^4}{4} + C_1\]\[u_2(x) = \int -x^3 dx = -\frac{x^4}{4} + C_2\]Since we need a particular solution, we can set the constants \( C_1 \) and \( C_2 \) to zero.

Step 4: Determine Particular Solution \(y_p(x)\)

Substitute \( u_1(x) \) and \( u_2(x) \) back into the expression for \( y_p(x) \):\[y_p(x) = \frac{x^4}{4} \cdot x + \left(-\frac{x^4}{4}\right) \cdot x^{-2} = \frac{x^5}{4} - \frac{x^2}{4}\]Simplify to obtain the particular solution:\[y_p(x) = \frac{x^5 - x^2}{4}\].

Step 5: Final Solution Statement

The particular solution to the given nonhomogeneous equation is:\[ y_p(x) = \frac{x^5 - x^2}{4} \] which, when added to the general solution of the homogeneous equation, provides the complete solution.

Key Concepts

Linear Differential Equations

A linear differential equation is an equation that involves a function and its derivatives. These are called 'linear' because they only involve linear terms in the function and its derivatives, meaning there are no powers or products of the function and its derivatives higher than the first degree. Such equations can be written in the general form:\[a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \ldots + a_1(x)y' + a_0(x)y = f(x)\]Here, \(a_n(x), a_{n-1}(x), \ldots, a_0(x)\) are coefficient functions of \(x\), and \(f(x)\) is a given function representing the nonhomogeneous part, or it may be zero for homogeneous equations. Solving these equations typically involve methods like undetermined coefficients, variation of parameters, or transform methods, depending on whether they are homogeneous or nonhomogeneous.

Particular Solution

When solving a differential equation, the particular solution represents one solution that satisfies the entire equation including the nonhomogeneous part. In the presence of a non-zero \(f(x)\), finding the particular solution involves matching the differential equation with this term.For instance, the method of variation of parameters helps to find a particular solution by constructing solutions based on the homogeneous solutions. Essentially, instead of an arbitrary combination, it involves specific modifications via auxiliary functions like \(u_1(x)\) and \(u_2(x)\) in\[y_p(x) = u_1(x)y_1(x) + u_2(x)y_2(x)\]where \(y_1(x)\) and \(y_2(x)\) are solutions to the corresponding homogeneous equation. This results in a solution that balances out the effect of the nonhomogeneous terms.

Wronskian

The Wronskian is a determinant used to study the linear independence of a set of solutions. It is particularly important when we try to find solutions to linear differential equations, especially when using methods like variation of parameters.To compute the Wronskian for two functions \(y_1(x)\) and \(y_2(x)\), it is given by:\[W = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} = y_1 \cdot y_2' - y_2 \cdot y_1'\]If the Wronskian is non-zero at any point in the interval where the solution is defined, the set of functions \(y_1(x)\) and \(y_2(x)\) are linearly independent, which is crucial for methods like variation of parameters to succeed. In our problem, it's calculated as -3x^{-2}, indicating that \(y_1\) and \(y_2\) are indeed linearly independent for \(x > 0\).

Nonhomogeneous Equations

A nonhomogeneous differential equation is one in which the function on the right-hand side, \(f(x)\), is not zero. It can be written as:\[a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \ldots + a_1(x)y' + a_0(x)y = f(x)\]The presence of \(f(x)\) sets these equations apart from homogeneous ones. Finding the solution usually involves finding a general solution to the corresponding homogeneous equation and a particular solution that accounts for the \(f(x)\) term.

• General Solution: Involves the complete set of solutions to the associated homogeneous equation.
• Particular Solution: Balances the nonhomogeneous term \(f(x)\).

The complete solution of a nonhomogeneous equation is then the sum of the general solution of the homogeneous equation and the particular solution of the nonhomogeneous equation.