91Ó°ÊÓ

The substitution method is a powerful tool for transforming complex differential equations into simpler forms. In the case of Cauchy-Euler equations, which typically involve terms with variable coefficients, a substitution such as \( x = e^t \) can transform the equation into one with constant coefficients. This simplifies solving the differential equation significantly.
In this context, substituting \( x = e^t \) alters the independent variable from \( x \) to \( t \). The derivatives with respect to \( x \) are recalculated in terms of \( t \). This makes the equation much easier to handle analytically.

• The first derivative transforms as \( y' = \frac{dy}{dt} \cdot \frac{dt}{dx} \), which simplifies to \( y'/x \).
• The second derivative becomes \( y'' = \frac{d^2y}{dt^2} \cdot \frac{1}{x^2} + \frac{dy}{dt} \cdot \frac{1}{x^2} \).

After substituting these transformed derivatives into the original equation, the complexity is reduced, enabling the use of techniques suited for constant coefficient differential equations.

Differential equations with constant coefficients are easier to solve compared to those with variable coefficients. Once a Cauchy-Euler equation is transformed into this form using substitution, the solution process involves finding the roots of a related characteristic equation.
The transformation leads to a linear differential equation such as \( 2 \frac{d^2y}{dt^2} - 5 \frac{dy}{dt} - 3y = 1 + 2 e^{t} + e^{2t} \). Here, the coefficients of the differential terms are constants, making it straightforward to solve.

• Write the characteristic equation based on the transformed DE’s homogeneous part, e.g., \( 2r^2 - 5r - 3 = 0 \).
• Solve this quadratic equation for \( r \) to find the roots \( r_1 \) and \( r_2 \).

These roots directly inform the structure of the homogeneous solution, a critical component in constructing the general solution.

The method of undetermined coefficients is a systematic way to find particular solutions to non-homogeneous linear differential equations. It is especially useful when the non-homogeneous part is a combination of polynomial, exponential, or trigonometric functions.
In the given equation, after transformation, the particular solution \( y_p(t) \) is assumed in a form like \( A + B e^t + C e^{2t} \) reflecting the nature of the non-homogeneous term \( 1 + 2 e^t + e^{2t} \).

• Substitute \( y_p(t) \) into the differential equation.
• Match coefficients on both sides of the equation to solve for constants \( A, B, \) and \( C \).

This step harmonizes the particular solution so it complements the homogeneous solution, thereby enabling the complete solution of the differential equation.

A homogeneous solution to a differential equation is part of the overall solution that solves the equation when there is no external forcing function (i.e., the non-homogeneous term is zero).
For the homogeneous part of the transformed equation \( 2 \frac{d^2y}{dt^2} - 5 \frac{dy}{dt} - 3y = 0 \), we use the roots of its characteristic equation \( 2r^2 - 5r - 3 = 0 \) to build the solution.

• If the roots \( r_1 \) and \( r_2 \) are distinct, the solution will be \( y_h(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \).
• If the roots are complex or repeated, the form of \( y_h(t) \) adjusts accordingly, using sines/cosines or polynomial terms.

This part of the solution captures the system's inherent dynamics without external influence.