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The characteristic equation is a powerful concept dealing with linear differential equations with constant coefficients. Whenever you are tasked with solving such differential equations, like the one given:

• The characteristic equation lets you deduce important properties to find solutions.
• For the differential equation \(6y'' - y' - y = 0\), the characteristic equation is obtained by replacing \(y''\) with \(r^2\), \(y'\) with \(r\), and \(y\) with \(1\), leading to:

\[ 6r^2 - r - 1 = 0 \]This converts the differential equation into a polynomial format, offering a simpler approach to tackling it. You then focus on solving this characteristic equation, which aligns with many algebraic techniques taught in earlier math courses.
As you see, understanding the formation of the characteristic equation is crucial. It provides a bridge between differential equations and algebra, simplifying the path to a solution.

The quadratic formula is a timeless tool in algebra used to solve quadratic equations. When your characteristic equation takes on the quadratic form \(ar^2 + br + c = 0\), like from above where \(a = 6\), \(b = -1\), \(c = -1\), the quadratic formula comes to the rescue:

• \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Applying it to our equation \(6r^2 - r - 1 = 0\), first compute the discriminant \((b^2 - 4ac)\):

• \((-1)^2 - 4(6)(-1) = 1 + 24 = 25\).
• The square root of 25 is 5, leading to:
• \( r = \frac{1 \pm 5}{12} \)

This simplifies to the roots \(r_1 = \frac{1}{2}\) and \(r_2 = -\frac{1}{3}\). Knowing how to effectively use the quadratic formula turns daunting quadratic equations into manageable problems, swiftly guiding you to the solutions.

The general solution is the ultimate goal in solving differential equations, representing a family of solutions that incorporate constants. Once you have the roots \(r_1 = \frac{1}{2}\) and \(r_2 = -\frac{1}{3}\) from your characteristic equation:

• You can formulate the general solution using these roots.
• Since the roots are real and distinct, the form is:

\[ y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \]Substituting \(r_1\) and \(r_2\):

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

The constants \(C_1\) and \(C_2\) represent any real numbers, determined by initial conditions or boundary conditions specific to a problem. This form of the general solution is pivotal for checking against specific conditions and ensuring the differential equation properly reflects the modeled system in practical applications. This understanding is key for deeper insights into the behavior of solutions across different scenarios.