91Ó°ÊÓ

The characteristic equation is a fundamental step in solving second order linear differential equations. It links the differential equation to a simpler polynomial equation which we can solve more easily. For a typical differential equation of the form \(ay'' + by' + cy = 0\), the characteristic equation is a polynomial given by \(ar^2 + br + c = 0\). This equation arises because we assume solutions of the form \(y(t) = e^{rt}\), where \(r\) is a constant. Substituting this assumption into the differential equation transforms it to the characteristic polynomial. This reduced, simpler form allows us to focus on finding \(r\), the roots, which are crucial in determining the nature of the solution. For our exercise, the differential equation \(3y'' + 11y' - 7y = 0\) translates to the characteristic equation \(3r^2 + 11r - 7 = 0\). Successfully solving it is the gateway to finding the complete solution of the original differential equation.

The quadratic formula is an essential tool for solving quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). In a second order differential equation, the characteristic equation often turns out to be a quadratic. To solve \(3r^2 + 11r - 7 = 0\) via the quadratic formula, recall the formula: \[r = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\]Here, the coefficients \(a=3\), \(b=11\), and \(c=-7\) need to be substituted into the formula. Following this:

• First, calculate the discriminant: \(b^2 - 4ac\)
• This will determine the nature of the roots: real or complex
• Then, solve for \(r\) to find both possible values \(r_1\) and \(r_2\)

In this exercise, inserting these values, we find: \[r = \frac{-11 \pm \sqrt{(11)^2 - 4 \times 3 \times (-7)}}{2 \times 3}\]. Solving this delivers the roots of the characteristic equation, pivotal for finding our final solution.

The general solution of a differential equation provides the family of all possible solutions, related to the arbitrary constants \(c_1\) and \(c_2\). In this context, it derives from the roots of the characteristic equation. Once the characteristic equation \(3r^2 + 11r - 7 = 0\) has been solved for roots \(r_1\) and \(r_2\), they dictate the form of the solution. For a second order linear differential equation with distinct real roots, the general solution is given by:\[y(t) = c_1e^{r_1 t} + c_2e^{r_2 t}\]Here:

• \(c_1\) and \(c_2\) are constants derived from initial conditions or specific criteria.
• \(e^{r_1 t}\) and \(e^{r_2 t}\) are the exponential solutions corresponding to each root.

Through this method, the solutions form a continuous spectrum of possibilities, all arising from different initial conditions or constraints applied to the original problem. Solving for \(c_1\) and \(c_2\) using particular values of \(y\) and its derivative allows us to find specific solutions.