Question

In each of Problems I through 8 find the general solution of the given differential equation. $$ y^{\prime \prime}+2 y^{\prime}-3 y=0 $$

Short answer

Answer: The general solution of the given differential equation is \(y(x) = c_1 e^{x} + c_2 e^{-3x}\), where \(c_1\) and \(c_2\) are arbitrary constants.

Step-by-step solution

Write the characteristic equation

To find the characteristic equation, we shall replace the terms involving derivatives with powers of a variable \(r\). So for the given equation: $$ y^{\prime \prime}+2 y^{\prime}-3 y=0 $$ The characteristic equation will be: $$ r^2 + 2r - 3 = 0 $$

Solve the characteristic equation

To find the roots of the characteristic equation, we can use the quadratic formula: $$ r = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$ For our equation, we have \(a = 1\), \(b = 2\), and \(c = -3\). Plugging these values into the formula gives: $$ r = \frac{-2 \pm \sqrt{2^2 - 4(1)(-3)}}{2(1)} \\ r = \frac{-2 \pm \sqrt{16}}{2} $$ This gives us two roots: \(r_1 = 1\) and \(r_2 = -3\).

Write the general solution

Since the roots are real and distinct, the general solution will be of the form: $$ y(x) = c_1 e^{r_1 x} + c_2 e^{r_2 x} $$ Using the roots found in step 2, the general solution is: $$ y(x) = c_1 e^{x} + c_2 e^{-3x} $$ This is the general solution of the given differential equation.

Key concepts

Characteristic Equation

The characteristic equation is a key aspect when solving linear differential equations with constant coefficients. It's essentially a bridge transferring a differential equation into an algebraic one, immensely simplifying the process of finding solutions.

To illustrate, let's say we have a differential equation such as the one in our exercise: \(y'' + 2y' - 3y = 0\). We assume solutions in the form of \(e^{rx}\), where \(r\) is what we need to find. By substituting derived forms of this assumed solution into the original differential equation, we can deduce a polynomial in terms of \(r\), and this polynomial is known as the characteristic equation. In our case, the characteristic equation is \(r^2 + 2r - 3 = 0\).

The roots of the characteristic equation provide us with the exponents for the terms in the general solution of the differential equation. By solving this equation, you are effectively finding the 'characteristics' of the solution to the original differential equation.

General Solution

The general solution of a differential equation encapsulates all possible specific solutions. It's a formula that includes arbitrary constants which can be determined if initial conditions are provided.

Our given differential equation has constant coefficients, and since we have distinct real roots from the characteristic equation, the general solution takes on a specific form: \(y(x) = c_1 e^{r_1 x} + c_2 e^{r_2 x}\), with \(c_1\) and \(c_2\) being the arbitrary constants and \(r_1\) and \(r_2\) the roots found previously.

Computing our roots as \(r_1 = 1\) and \(r_2 = -3\), we plug them into this general solution to get \(y(x) = c_1 e^{x} + c_2 e^{-3x}\). This beautifully shows that differential equations, quite intimidating at first glance, lead to an elegant summary of all possible behaviors of the system described by the equation—in this case, a weighted sum of exponential functions.

Quadratic Formula

The quadratic formula is the go-to tool for solving any quadratic equation, which is an equation of the form \(ax^2 + bx + c = 0\). The formula itself is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(\pm\) indicates that there are typically two solutions.

Applying it to our characteristic equation, we have \(a = 1\), \(b = 2\), and \(c = -3\). Plug those into the formula, and we find our roots, \(r_1 = 1\) and \(r_2 = -3\).

What's fantastic about the quadratic formula is its universality; it works for any quadratic equation. Remember, the discriminant \(b^2 - 4ac\) in the formula's radical decides the nature of the roots—whether they're real or complex, and distinct or repeated. The combination of roots informs the structure of the general solution for differential equations like the one we're exploring, showcasing the profound interconnectedness between algebra and calculus.