91Ó°ÊÓ

A characteristic equation plays a crucial role when solving a second-order linear homogeneous differential equation with constant coefficients. It is a bridge that connects our differential equation to algebra, making the solving process familiar and manageable. In this context, the differential equation \( \frac{d^{2} y}{d t^{2}} - 4 \frac{d y}{d t} - 5 y = 0 \) translates into the characteristic equation \( r^2 - 4r - 5 = 0 \). Here:

• \(a = 1\)
• \(b = -4\)
• \(c = -5\)

The solution to this characteristic equation will reveal the roots \( r \) which are essential to forming the general solution of the differential equation. Understanding how to form and solve the characteristic equation allows us to unlock solutions to many differential equations that follow a similar form.

An initial-value problem involves finding a solution to a differential equation that also satisfies specific conditions at a given point. Here, these conditions are specified as \( y(1) = 0 \) and \( y'(1) = 2 \).

This means that the solution to the differential equation must not only be general but also precise enough to meet these requirements:

• At \( t = 1, \) the function \( y(t) \) must equal 0.
• The derivative \( y'(t) \) must equal 2 at the same point \( t = 1 \).

These initial conditions provide the necessary constraints to determine the specific values of constants in the general solution, ensuring that the solution is unique to this particular problem.

The quadratic formula is a powerful tool for solving quadratic equations like the characteristic equation \( r^2 - 4r - 5 = 0 \). It is expressed as:\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]By identifying the coefficients:

• \( a = 1 \)
• \( b = -4 \)
• \( c = -5 \)

we substitute them into the formula and solve for \( r \):\[ r = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm 6}{2}\]resulting in the roots \( r = 5 \) and \( r = -1 \). With these roots, we can build the expression for the general solution of the differential equation. Knowing how to apply the quadratic formula is essential for efficiently tackling characteristic equations.

In solving a differential equation, the general solution is a comprehensive expression that captures all potential solutions. Once we have the roots \( r = 5 \) and \( r = -1 \) from the characteristic equation, we can write the general solution as:\[ y(t) = C_1 e^{5t} + C_2 e^{-t}\]where \(C_1\) and \(C_2\) are arbitrary constants. In this solution, each term corresponds to a root of the characteristic equation and represents a fundamental solution.The next step is to apply the initial conditions (from the initial-value problem) to this general form. By substituting the specific values of \( t \) into the general solution and its derivative, we derive a system of equations to solve for \( C_1 \) and \( C_2 \). The result is a particular solution tailored to the initial-value problem, transforming a general solution into a specific one relevant to the conditions given.