91Ó°ÊÓ

Differential equations are a crucial aspect of mathematics with numerous practical applications. A special kind of these is the second-order linear homogeneous equation. The differential equation provided in the exercise is of this type. This type is defined by the general structure:

• The highest derivative present is the second derivative, commonly denoted as \( y'' \).
• It is linear, meaning terms involving the function \( y \), and its derivatives are summed without any products or powers of these terms.
• It's homogeneous, suggesting that all terms are proportional to \( y \) or its derivatives, with no standalone functions of \( x \).

Understanding these properties aids in finding the solution, as they guide the inherent mathematical structure that can be leveraged. For instance, knowing that it's homogeneous implies that the solution can often be discovered using trial functions like \( y = x^m \).

In second-order linear homogeneous differential equations, coefficients can either be constant or variable. In this case, the given differential equation \( x^2 y'' - 3x y' + 4y = 0 \) has variable coefficients, evident because terms like \( x^2 \) and \( -3x \) multiply the derivatives.These coefficients affect the solution method. Unlike constant-coefficient equations where simpler algebraic techniques like the characteristic equation can be immediately applied, variable coefficient equations typically require a different approach. A common technique is the power law assumption, such as \( y = x^m \), which substitutes back into the equation and helps reduce it to an algebraic form that can be solved more directly.

Initial conditions provide crucial additional information that enables a unique solution to the differential equation. In this scenario, we are given \( y(1) = 5 \) and \( y'(1) = 3 \). These conditions specify the value of the function \( y \) and its first derivative at a particular point, here \( x = 1 \).These conditions tailor the general solution to meet specific constraints or scenarios, converting it into a particular solution. By applying the initial conditions, constants in the general solution, such as \( C_1 \) and \( C_2 \), can effectively be calculated to ensure the solution aligns perfectly with these prescribed values.

For differential equations with constant coefficients, the characteristic equation is a vital tool for finding solutions. However, with variable coefficients, as in our example equation \( x^2 y'' - 3x y' + 4y = 0 \), the method adapts slightly. Assume a solution of the form \( y = x^m \). When substituted into the equation, this assumption transforms the differential equation into an algebraic form.In this exercise, the assumed trial solution leads us to an equation in terms of \( m \):\[ m(m-1)x^m - 3mx^m + 4x^m = 0 \]This is simplified to the characteristic equation:\[ m^2 - 4m + 4 = 0 \]Solving this quadratic equation for \( m \) gives the roots necessary for constructing the general solution. In this case, the roots were a repeated root, \( m = 2 \), indicating that the general solution will include both a term involving \( x^m \) and \( x^m \ln(x) \) to account for the multiplicity of the root, resulting in a more complex solution structure.