91Ó°ÊÓ

The first step in solving a linear differential equation like \( y'' - y' + 2.5y = 0 \) is to rewrite it as a characteristic equation. Think of this process as converting the differentiation operation into an algebraic problem. The structure \( ay'' + by' + cy = 0 \) gives rise to the characteristic quadratic equation \( ar^2 + br + c = 0 \). This simplifies our work, as we can now solve for \( r \) rather than solving the differential equation directly.

For the given equation, the coefficients are \( a = 1 \), \( b = -1 \), and \( c = 2.5 \). Plugging these into the form, we generate \( r^2 - r + 2.5 = 0 \). In essence, the characteristic equation serves as a bridge, transforming a differential equation of functions into a simpler polynomial equation of variables.

The next step involves solving the characteristic equation, \( r^2 - r + 2.5 = 0 \), by finding its roots. To determine the nature of these roots, calculate the discriminant \( \Delta = b^2 - 4ac \). Here, it's calculated as \( (-1)^2 - 4(1)(2.5) = 1 - 10 = -9 \). A negative discriminant indicates the roots are complex numbers. This reveals the oscillator nature of the solution.

Use the quadratic formula \( r = \frac{-b \pm \sqrt{\Delta}}{2a} \) to find the roots. With \( \Delta = -9 \) and given \( a = 1 \), \( b = -1 \), we find the roots \( r = \frac{1 \pm 3i}{2} \). They are \( r_1 = \frac{1}{2} + \frac{3}{2}i \) and \( r_2 = \frac{1}{2} - \frac{3}{2}i \).

Complex roots have a real part and an imaginary part, expressed as \( r = \alpha \pm \beta i \), where \( \alpha \) affects exponential growth/decay and \( \beta \) describes oscillation.

Once you have complex roots, you can construct the general solution of the differential equation. For a pair of complex roots \( \alpha \pm \beta i \), the associated general solution is:

• \( y(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t)) \)

This form combines exponential and trigonometric functions, embodying both growth/decay and oscillatory behavior.

For our problem, \( \alpha = \frac{1}{2} \) and \( \beta = \frac{3}{2} \). Therefore, the solution is:

• \( y(t) = e^{\frac{1}{2}t}(C_1 \cos(\frac{3}{2}t) + C_2 \sin(\frac{3}{2}t)) \)

Here, \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions. This general solution incorporates the essence of the system's dynamics, allowing us to predict the behavior for different initial values.