91Ó°ÊÓ

To verify a solution of a differential equation, you need to check if the solution satisfies the equation when substituted back in. Verification can confirm that a given solution, like \( y_1 = x \), is indeed a solution to the homogeneous form of the differential equation.
We start by recognizing the homogeneous equation derived from the original: \[ x^2 y'' - (x^2 + 2x) y' + (x+2)y = 0 \].Here are the steps:

• Calculate the first and second derivatives of \( y_1 = x \).
• Substitute \( y_1 \), \( y_1' \), and \( y_1'' \) into the differential equation.
• Simplify the resulting expression.

For \( y_1 = x \), \( y_1' = 1 \) and \( y_1'' = 0 \). Substitution yields:\[ x^2(0) - (x^2 + 2x)(1) + (x+2)(x) = 0 \]Simplified, this becomes:\[ -x^2 - 2x + x^2 + 2x = 0 \]The equation holds true, so \( y_1 = x \) is a verified solution of the homogeneous equation.

The method of reduction of order allows you to find a second independent solution of a second-order linear homogeneous differential equation when one solution is already known. This is useful when you need to determine a complete set of solutions for a differential equation.In our case, we have \( y_1 = x \) as a known solution. We assume the second solution \( y_2 \) to be of the form:\[ y_2 = v(x) y_1 = v(x) x \]Steps involved in reduction of order:

• Express \( y_2' \) and \( y_2'' \) using the product rule.


• Substitute \( y_2 \), \( y_2' \), \( y_2'' \) back into the homogeneous differential equation.
• Simplify to obtain an equation involving \( v' \).

After substituting, the equation simplifies to:\[ x^2v'' + 2xv' = 0 \]By making a substitution \( v' = u \) and solving for \( v \), we find that the second solution is \( y_2 = x \ln(x) \).This derived solution \( y_2 \) is linearly independent from \( y_1 \), expanding the solution space.

A homogeneous equation in the context of differential equations means every term is reliant on the function and its derivatives alone, with no external non-zero terms. It effectively looks like this:\[ a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \, ... \, + a_0(x)y = 0 \]In our exercise, the differential equation \[ x^2 y'' - (x^2 + 2x) y' + (x+2)y = 0 \]is our homogeneous equation.Key Characteristics:

• It only includes terms of the dependent variable (in this case, \( y \)) and its derivatives.


• The absence of any standalone terms (like constants or functions not multiplied by \( y \)).
• Any solution to this equation only requires combinations of derivative terms to total zero.

Understanding the structure of a homogeneous equation is crucial because the solutions give us the base upon which we can build a complete solution to a related nonhomogeneous equation.

When dealing with a nonhomogeneous differential equation, such as:\[ x^2 y'' - (x^2 + 2x) y' + (x+2)y = x^3 \],you need to find a particular solution, \( y_p \), that satisfies the entire equation. This solution accounts for the non-zero terms on the right-hand side of the equation.Finding a Particular Solution:

• The method of undetermined coefficients or variation of parameters can be applied.


• For our exercise, the particular solution was found through variation of parameters using the known solutions \( y_1 = x \) and \( y_2 = x \ln(x) \).
• The result was the particular solution \( y_p = \frac{x^3}{6} \).

Combining \( y_p \) with the solutions of the homogeneous equation (\( y_1 \) and \( y_2 \)), forms the general solution:\[ y(x) = C_1 x + C_2 x \ln(x) + \frac{x^3}{6} \].The particular solution ensures the full equation is satisfied, reflecting both the homogeneous structure and nonhomogeneous features.