Problem 47 - Calculus Volume 3

Question

Find a differential equation with a general solution that is \(y=c_{1} e^{x / 5}+c_{2} e^{-4 x}\).

Step-by-Step Solution

Answer

The differential equation is \( \frac{d^2 y}{dx^2} + \frac{19}{5} \frac{dy}{dx} - \frac{4}{5} y = 0 \).

Step 1: Identify General Solution Form

We start with the given general solution: \[y = c_{1} e^{x/5} + c_{2} e^{-4x}\]This is a linear combination of exponential functions, which typically corresponds to a second-order linear differential equation with constant coefficients.

Step 2: Write the Characteristic Equation

The solution consists of exponential terms, which implies the differential equation has a characteristic equation. The characteristic equation has roots corresponding to the exponents of the exponential functions: \[r_{1} = \frac{1}{5}, \quad r_{2} = -4\]Thus, the characteristic equation is:\[(r - \frac{1}{5})(r + 4) = 0\]Expanding this, we get:\[r^2 + \frac{19}{5}r - \frac{4}{5} = 0\]

Step 3: Formulate the Differential Equation

The roots of the characteristic equation correspond to a quadratic form obtained from the differential equation:\[a(D) = D^2 + \frac{19}{5}D - \frac{4}{5}\]where \( D \) denotes the derivative operator \( \frac{d}{dx} \). Therefore, the differential equation is:\[D^2 y + \frac{19}{5} D y - \frac{4}{5} y = 0\]or, written explicitly in terms of derivatives:\[\frac{d^2 y}{dx^2} + \frac{19}{5} \frac{dy}{dx} - \frac{4}{5} y = 0\]

Step 4: Verify the Solution

We can verify the solution by substituting back into the differential equation:1. Calculate the first and second derivatives: - \( y = c_{1} e^{x/5} + c_{2} e^{-4x} \) - \( y' = \frac{c_{1}}{5} e^{x/5} - 4c_{2} e^{-4x} \) - \( y'' = \frac{c_{1}}{25} e^{x/5} + 16c_{2} e^{-4x} \)2. Substituting into the differential equation: - \( y'' + \frac{19}{5}y' - \frac{4}{5}y = 0 \)3. Simplify to confirm zero results, ensuring it is correct.

Key Concepts

Second-Order Differential Equations

A second-order differential equation is an equation that involves the second derivative of a function. In simple terms, it's like writing an equation that links a function with both its rate of change (its first derivative) and the change in that rate (its second derivative). These equations are called "second-order" because the highest derivative is the second one.

Second-order differential equations are essential because they describe many physical phenomena, such as how objects move or how heat flows.

• The standard form is usually written as \(a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x)\), where \(a\), \(b\), and \(c\) are constants, and \(f(x)\) is a given function.
• If \(f(x)\) is zero, it's called a "homogeneous" equation.
• They often involve solving with initial conditions to find a particular solution.

Understanding these concepts helps you to unravel complex relationships within different scientific fields.

Characteristic Equation

The characteristic equation is a crucial concept when dealing with second-order differential equations, especially those with constant coefficients. It is derived from the original differential equation and gives insights into the nature of the solutions.

When you have a homogeneous second-order differential equation with constant coefficients, the characteristic equation can be formed by replacing each derivative with a power of \(r\), like \(r^2\) for \(\frac{d^2}{dx^2}\):

• This transformation turns the differential equation into a characteristic polynomial equation.
• In our example, this resulted in \((r - \frac{1}{5})(r + 4) = 0\).
• The roots of this polynomial (\(r_1 = \frac{1}{5}\) and \(r_2 = -4\)) are vital as they directly point towards the solution of the original differential equation.

The characteristic equation acts as a bridge from a differential problem to a polynomial solution, helping you find solutions more efficiently.

General Solution

The general solution of a differential equation is a solution that incorporates all possible solutions for the equation in a broad sense. This entails the complete set of functions that satisfy the differential equation.

Especially in the context of second-order differential equations, the general solution frequently includes two arbitrary constants, like \(c_1\) and \(c_2\). These constants represent the degree of freedom due to the order of the differential equation:

• Our example gives us a general solution in terms of exponential functions, \(y = c_{1} e^{x/5} + c_{2} e^{-4x}\).
• The arbitrary constants \(c_1\) and \(c_2\) allow the solution to fit a wide range of initial or boundary conditions.
• The presence of two independent solutions, \(e^{x/5}\) and \(e^{-4x}\), shows that it covers the entire solution space.

In practice, you’ll need specific conditions, known as initial conditions, to pinpoint a unique solution from this general one.

Exponential Functions

Exponential functions play a significant role in the solutions of differential equations. They are functions in the form \(e^{kx}\), where \(e\) is the base of natural logarithms and \(k\) is any real number.

These functions are incredibly powerful in representing growth or decay processes and appear prominently in the solutions of linear differential equations with constant coefficients:

• They arise naturally in scenarios involving repeated processes or equal interval changes, such as populations growth or radioactive decay.
• In our solutions, \(e^{x/5}\) and \(e^{-4x}\) reflect typical solution structures for the given differential equation's roots.
• The differing values for \(k\) (positive or negative) indicate different types of behavior, like growing or shrinking exponentially.

Expanding your understanding of exponential functions will aid you in interpreting and solving various kinds of differential equations effectively.