91Ó°ÊÓ

When dealing with second-order linear homogeneous differential equations, an essential step is finding the characteristic equation. This equation paves the journey to solving the differential equation. For an equation like \(y'' + 6y' + 8y = 0\), we begin by assuming a solution of the form \(y = e^{rt}\). This assumption is helpful because it converts the differentiation process into a simpler algebraic form. Here's why:

• When we differentiate \(y = e^{rt}\), we get \(y' = re^{rt}\) and \(y'' = r^2e^{rt}\).
• Plugging these back into the differential equation replaces the derivatives with powers of \(r\), producing the characteristic equation \(r^2 + 6r + 8 = 0\).

This characteristic equation is a crucial step because solving it gives us the values of \(r\) that satisfy the differential equation, thereby leading us towards the general solution.

The general solution to a differential equation like \(y^{\prime\prime} + 6y^{\prime} + 8y = 0\) depends significantly on the roots of its characteristic equation. In this case, the characteristic equation is \(r^2 + 6r + 8 = 0\).Once we find the solutions (or roots) to this equation, we can construct the general solution. For distinct real roots, \(r_1\) and \(r_2\), the general solution takes the form:

• \(y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}\),

where \(C_1\) and \(C_2\) are constants determined later by initial conditions. In our example, we have identified the roots \(r_1 = -2\) and \(r_2 = -4\), leading to the general solution:\[ y(t) = C_1 e^{-2t} + C_2 e^{-4t}. \]This expression captures all possible solutions to the differential equation, providing a complete description in terms of \(t\). It’s a powerful portrayal because it fits infinite scenarios by adjusting \(C_1\) and \(C_2\).

The quadratic formula is a fundamental mathematical tool used to solve quadratic equations of the form \(ax^2 + bx + c = 0\). When the characteristic equation \(r^2 + 6r + 8 = 0\) emerges from our differential scenario, the quadratic formula comes into play.The quadratic formula is given as:\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]where \(a\), \(b\), and \(c\) represent the coefficients from the quadratic equation. Applying it to our characteristic equation:

• \(a = 1\),
• \(b = 6\),
• \(c = 8\).

Plugging these values into the quadratic formula, we compute:\[ r = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}. \]This simplifies as:

• \[ r = \frac{-6 \pm \sqrt{36 - 32}}{2} = \frac{-6 \pm 2}{2}, \]

which results in roots \(r_1 = -2\) and \(r_2 = -4\). The quadratic formula not only helps solve numbers; it's a bridge connecting the algebra of differential equations to their most elegant solutions.