91Ó°ÊÓ

Once we identify a second-order linear homogeneous differential equation with constant coefficients, such as:

• \(9y'' + 12y' + 4y = 0\),

we need to form what's called the characteristic equation. This helps transform the differential equation into a simpler algebraic form. To write it, we use a substitute: if the differential equation is:

• \(ay'' + by' + cy = 0\)

its characteristic equation is:

• \(ar^2 + br + c = 0\).

The mechanism behind this is substituting potential solutions of the form \(e^{rx}\), due to its predictive growth nature when multiplied by its derivatives. It simplifies understanding how solutions behave. When we plug in the given values: \(a = 9\), \(b = 12\), and \(c = 4\), we derive the equation:\[9r^2 + 12r + 4 = 0\].This equation is essential for determining the type of solutions our differential equation will have.

A homogeneous differential equation is a type that is defined by this feature: every term is a multiple of the dependent variable or its derivatives. This means that if all terms can be "removed" by setting the variable to zero, we have a homogeneous equation.For example, in our differential equation:

• \(9y'' + 12y' + 4y = 0\),

the right-hand side is zero, making it homogeneous. Formally, an equation of the form

• \(ay'' + by' + cy = 0\)

is homogeneous. Since it's homogeneous, this equation ensures that if \(y = 0\), then all terms lead to zero, affirming its characteristic. This classification is crucial because solutions can be constructed based solely on the system's inherent characteristics without external inputs or forcing functions.

Once the characteristic equation is established and solved, findings may indicate various root scenarios. In our case, solving:\[9r^2 + 12r + 4 = 0\]using the quadratic formula gives a discriminant (\(b^2 - 4ac\)) of 0. This points towards repeated roots, which means there is only one single root repeated twice. When calculated:\[r = \frac{-12}{18} = -\frac{2}{3}\]it shows a double root of \(-\frac{2}{3}\).Having repeated roots in practicality leads to a solution form different from distinct roots. Instead of two separate exponential solutions, the general solution becomes:

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

where \(C_1\) and \(C_2\) are constants. This form arises because a single exponential does not span the solution space adequately compared to double distinct solutions, so multiplication by \(x\) compensates for the repeated effect, still satisfying the differential setup.