Problem 85 - 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. \(x^{2} y^{\prime \prime}-2 y=10 x^{2}-1\), \(y_{1}(x)=x^{2}\), \(y_{2}(x)=x^{-1}\)

Step-by-Step Solution

Answer

The particular solution is \( y_p = \frac{x^4}{3} - \frac{x^2 \ln|x|}{3} + \frac{x^2}{9} \).

Step 1: Verify the given solutions

Verify that the given solutions \( y_1 = x^2 \) and \( y_2 = x^{-1} \) are indeed solutions to the homogeneous equation. For this, substitute them individually into the homogeneous equation \( x^2 y'' - 2y = 0 \) and confirm both satisfy it.

Step 2: Write the general solution

The general solution for the homogeneous equation is \( y_h = C_1 y_1 + C_2 y_2 = C_1 x^2 + C_2 x^{-1} \) where \( C_1 \) and \( C_2 \) are constants.

Step 3: Solve the Wronskian

Calculate the Wronskian \( W(x) = y_1 y_2' - y_2 y_1' \):- Compute derivatives: \( y_1' = 2x \), \( y_2' = -x^{-2} \).- Substitute into the Wronskian formula: \( W(x) = x^2 (-x^{-2}) - x^{-1} (2x) = -1 - 2 = -3 \).

Step 4: Write integral equations for the particular solution

Using variation of parameters, we have two functions \( u_1 \) and \( u_2 \):\[ u_1(x) = -\int \frac{y_2(x) \cdot g(x)}{W(x)} \, dx = -\int \frac{x^{-1} (10x^2 - 1)}{-3} \, dx \].\[ u_2(x) = \int \frac{y_1(x) \cdot g(x)}{W(x)} \, dx = \int \frac{x^2 (10x^2 - 1)}{-3} \, dx \].Where \( g(x) = 10x^2 - 1 \).

Step 5: Solve for the integrals

Substitute and solve the integrals:- For \( u_1(x) \): \[ u_1(x) = \frac{1}{3} \int (10x - x^{-1}) \, dx = \frac{1}{3} (5x^2 - \ln|x|) \].- For \( u_2(x) \): \[ u_2(x) = -\frac{1}{3} \int (10x^4 - x^2) \, dx = -\frac{1}{3} \left(2x^5 - \frac{x^3}{3}\right) \].

Step 6: Formulate the particular solution

Thus, the particular solution \( y_p \) is:\[ y_p = u_1 y_1 + u_2 y_2 = \left(\frac{5x^2 - \ln|x|}{3}\right)x^2 - \left(\frac{2x^5}{3} - \frac{x^3}{9}\right)x^{-1} \].

Step 7: Simplify the particular solution

Simplify \( y_p \):\[ \Rightarrow y_p = \frac{5x^4}{3} - \frac{x^2 \ln|x|}{3} - \frac{2x^4}{3} + \frac{x^2}{9} \].Combine like terms: \[ y_p = \frac{x^4}{3} - \frac{x^2 \ln|x|}{3} + \frac{x^2}{9} \].

Key Concepts

Linearly Independent Solutions

In differential equations, it is crucial to identify linearly independent solutions for a homogeneous equation. Linearly independent functions are those that cannot be expressed as a scalar multiple of one another. This property ensures that the solutions capture all possible behaviors of the differential equation.

For the given equation \( x^{2} y^{\prime \prime}-2 y=0 \), the solutions \( y_1(x)=x^2 \) and \( y_2(x)=x^{-1} \) are linearly independent. How do we know this? Because neither of these solutions can be expressed as a simple multiple of the other. Linearly independent solutions form the foundation of the general solution of a homogeneous differential equation.

When constructing the general solution \( y_h \), we use these linearly independent solutions to frame it as \( y_h = C_1 y_1 + C_2 y_2 = C_1x^2 + C_2x^{-1} \), where \( C_1 \) and \( C_2 \) are constants. This method helps us capture the full family of solutions of the homogeneous equation.

Wronskian

The Wronskian is a determinant used to determine whether a set of solutions is linearly independent. For functions \( y_1 \) and \( y_2 \), the Wronskian \( W(x) \) is given by the formula: \[ W(x) = y_1 y_2' - y_2 y_1' \]

In our exercise, with \( y_1(x) = x^2 \) and \( y_2(x) = x^{-1} \), this becomes:

• Calculate the derivatives: \( y_1' = 2x \) and \( y_2' = -x^{-2} \).
• Substitute into the Wronskian formula: \( W(x) = x^2(-x^{-2}) - x^{-1}(2x) \).
• Simplify: \( W(x) = -1 - 2 = -3 \).

A non-zero Wronskian, such as \(-3\) in this case, confirms that the solutions \( y_1 \) and \( y_2 \) are indeed linearly independent. This shows the solutions are suitable for constructing the general solution of the differential equation.

Nonhomogeneous Differential Equation

A nonhomogeneous differential equation is one where a term exists independently from the dependent variable and its derivatives. In contrast to a homogeneous equation, it has a non-zero term on one side, like \( 10x^2 - 1 \) in our equation \( x^{2} y^{\prime \prime}-2 y=10 x^{2}-1 \).

To find a particular solution, the method of variation of parameters is employed. This involves modifying the general solution of the homogeneous equation by adding a particular solution to account for the nonhomogeneous part. In this exercise, we specifically form integral equations using the functions \( u_1(x) \) and \( u_2(x) \) to determine this particular solution.

These functions are found using integrals:

• For \( u_1(x) \): Solve \(-\int \frac{x^{-1} (10x^2 - 1)}{-3} \, dx \).
• For \( u_2(x) \): Solve \( \int \frac{x^2 (10x^2 - 1)}{-3} \, dx \).

Solving these integrals gives us explicit expressions for \( u_1(x) \) and \( u_2(x) \), which ultimately help in crafting the particular solution \( y_p \).

This method ensures we incorporate both the homogeneous and nonhomogeneous parts in our overall solution \( y = y_h + y_p \).