Question

Solve the following differential equations. \(D(D+5) y=0\)

Short answer

The solution is \(y(x) = C_1 + C_2 e^{-5x}\).

Step-by-step solution

Rewrite the differential equation

The given differential equation is in terms of the operator notation. Rewrite the equation: \(D(D+5) y=0\).This can be expanded to: \(D^2 y + 5 D y = 0\).

Solve the characteristic equation

Transform the differential equation into its characteristic equation. For the operator \(D\), where \(D = \frac{d}{dx}\), the characteristic equation is given by:\(r^2 + 5r = 0\).

Find the roots

Solve the characteristic equation for \(r\): \(r^2 + 5r = 0\). This can be factored as: \(r(r + 5) = 0\). Thus, the roots are: \(r = 0\) and \(r = -5\).

Write the general solution

With the roots \(r_1 = 0\) and \(r_2 = -5\), the general solution to the differential equation is: \(y(x) = C_1 e^{0x} + C_2 e^{-5x}\).Simplify this to:\(y(x) = C_1 + C_2 e^{-5x}\), where \(C_1\) and \(C_2\) are constants.

Key concepts

Characteristic Equation

When solving a differential equation, transforming it into its characteristic equation is a crucial step. The characteristic equation serves as a bridge between the differential equation and algebra, allowing us to find solutions more easily.
For our given differential equation: \(D(D+5) y=0\). We convert this into \(r^2 + 5r = 0\), by substituting each differential operator \(D\) with \(r\). This makes the equation algebraic, easier to solve.
In general, converting to a characteristic equation works for linear differential equations with constant coefficients. Here are the steps summarized:

• Replace each differential operator with a variable, usually denoted as \(r\).
• Set up an algebraic equation.
• Solve the algebraic equation to find the roots.

This transformation helps to find the solutions, called roots, which are essential for the next steps.

General Solution

The general solution of a differential equation represents all potential solutions, encapsulating all possible behaviors of the equation.
For our example, after finding the roots of the characteristic equation \(r = 0\) and \(r = -5\), we use them to construct the general solution:

• For each root, we write a term \(C_i e^{r_i x}\), where \(C_i\) is a constant and \(r_i\) is a root.
• Here we get two terms: \(C_1 e^{0x}\) and \(C_2 e^{-5x}\).
• Summing these terms yields the general solution: \(y(x) = C_1 + C_2 e^{-5x}\).

The general solution includes all specific solutions by varying the constants \(C_1\) and \(C_2\). This represents multiple behaviors based on initial conditions.

Roots of Polynomial

The roots of a polynomial are the values that satisfy the equation when it is set to zero. In our characteristic equation \(r^2 + 5r = 0\), the roots are solutions to this equation.
To find the roots:

• First, factor the polynomial: \(r(r + 5) = 0\).
• Next, set each factor to zero: \(r = 0\) and \(r + 5 = 0\).
• Solve each factor: \(r = 0\) and \(r = -5\).

These are the roots of our characteristic polynomial. Each root represents an exponential solution to the original differential equation. Hence, finding these roots is pivotal for forming the general solution.