Question

Solve the following differential equations. \(\left(2 D^{2}+D-1\right) y=0\)

Short answer

The general solution is \(y(x) = C_1 e^{\frac{1}{2}x} + C_2 e^{-x}\).

Step-by-step solution

Identify the differential equation

The given differential equation is \((2D^2 + D - 1)y = 0\), where \(D\) represents the differentiation operator \(\frac{d}{dx}\).

Write down the characteristic equation

To solve the differential equation, we first need to find the characteristic equation. For the given differential operator \(2D^2 + D - 1\), the characteristic equation is formed by replacing \(D\) with \(r\): \[2r^2 + r - 1 = 0\]

Solve the characteristic equation

Solve the characteristic equation \(2r^2 + r - 1 = 0\) using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 1\), and \(c = -1\): \[r = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}\]This gives us the roots: \[r_1 = \frac{-1 + 3}{4} = \frac{1}{2}\] \[r_2 = \frac{-1 - 3}{4} = -1\]

Write the general solution

Since the characteristic equation has distinct real roots \(r_1 = \frac{1}{2}\) and \(r_2 = -1\), the general solution to the differential equation is: \[y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}\] Substituting the values of \(r_1\) and \(r_2\): \[y(x) = C_1 e^{\frac{1}{2}x} + C_2 e^{-x}\]

Key concepts

Differential Equations

A differential equation is a mathematical equation that relates a function with its derivatives. In simpler terms, it specifies how a function changes over time or space.

These equations are key in various fields like physics, engineering, and economics because they describe the behavior of systems. For example, they can model how populations grow, how heat spreads, or how electrical circuits operate.

In our exercise, we have the differential equation \(\left(2 D^{2}+D-1\right) y=0\), where \(D\) represents the differentiation operator \(\frac{d}{dx}\). This equation is also known as a linear homogeneous differential equation because it has constant coefficients and equates to zero.

Solving this type of equation often involves converting it into a simpler algebraic form, which leads us to the next concept: the characteristic equation.

Characteristic Equation

The characteristic equation is a polynomial equation derived from a differential equation.

To obtain this, we replace the differentiation operator \(D\) by a variable, usually denoted by \(r\). In our case, the differential operator is \(2D^2 + D - 1\), so the characteristic equation becomes \(2r^2 + r - 1 = 0\).

The characteristic equation simplifies the problem by transforming a differential equation into an algebraic one. Solving this polynomial equation gives us the values of \(r\), often called roots, which are essential for finding the general solution.

Quadratic Formula

The quadratic formula is a standard method to solve second-degree polynomial equations of the form \(ax^2 + bx + c = 0\).

The formula is given by: \[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].

In our example, the characteristic equation \(2r^2 + r - 1 = 0\) is a second-degree polynomial where \(a = 2\), \(b = 1\), and \(c = -1\).

Plugging these values into the quadratic formula, we get:

\[r = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}\].

This results in two roots: \(r_1 = \frac{1}{2}\) and \(r_2 = -1\).

General Solution

The general solution of a differential equation combines all individual solutions using arbitrary constants.

Once we have the roots of the characteristic equation, we can express the solution in terms of these roots.

For distinct real roots like \(r_1\) and \(r_2\), the general solution takes the form: \[y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}\].

Substituting the roots \(r_1 = \frac{1}{2} \) and \(r_2 = -1 \), we find: \[y(x) = C_1 e^{\frac{1}{2}x} + C_2 e^{-x}\].

Here, \(C_1\) and \(C_2\) are arbitrary constants determined by initial conditions or boundary conditions provided in specific problems.