91Ó°ÊÓ

In the study of ordinary differential equations (ODEs), particularly linear differential equations with constant coefficients, the characteristic equation plays a crucial role. To derive the characteristic equation, we convert the differential equation into a simpler algebraic form. This is done by assuming solutions of a specific type, typically exponential functions. Consider the initial differential equation given as:

\(6 y^{\prime \prime}+y^{\prime}-2 y=0\)

Here, the coefficients \(a\), \(b\), and \(c\) relate to the second derivative \(y''\), the first derivative \(y'\), and the function \(y\) itself.

• Second derivative coefficient: \(a = 6\)
• First derivative coefficient: \(b = 1\)
• Function coefficient: \(c = -2\)

The characteristic equation derived from this differential equation is:
\[ ar^2 + br + c = 0 \]
Which gives us:
\[ 6r^2 + r - 2 = 0 \]
This quadratic equation represents the characteristic equation whose roots determine the behavior of the solutions to the differential equation.

Quadratic roots are solutions to the quadratic equation, which is often expressed in the form:
\[ ax^2 + bx + c = 0 \]
Utilizing the quadratic formula can help us find these roots efficiently:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our provided characteristic equation \(6r^2 + r - 2 = 0\), we can apply these steps to find the roots:

• Calculate the discriminant: \(b^2 - 4ac = 1^2 - 4(6)(-2) = 1 + 48 = 49\)
• Substitute the values into the quadratic formula:

\[ r = \frac{-1 \pm \sqrt{49}}{12} \]

• Simplify to get the roots \(r_1 = \frac{1}{3}\) and \(r_2 = -1\).

These roots indicate whether the solutions will be real or complex, influencing the form of the general solution for the differential equation.

The general solution of a differential equation involves using the roots of the characteristic equation. This construction offers insight into how solutions behave over time.
Our roots, \(r_1 = \frac{1}{3}\) and \(r_2 = -1\), are both real and distinct. This means the general solution follows a specific structure:
\[ y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t} \]
In our case,
\[ y(t) = c_1 e^{\frac{1}{3} t} + c_2 e^{-t} \]
Where \(c_1\) and \(c_2\) are arbitrary constants determined by the initial conditions of a specific problem.

• If the roots were complex, the solution would involve sine and cosine functions instead.
• If the roots were equal, the solution form would include \(t\) as a factor, incorporating polynomial terms.

Understanding the nature of the roots is pivotal in crafting the final form of the general solution to any differential equation.