Problem 7 - Differential Equations and Linear Algebra

Question

Verify that the given function is a solution to the given differential equation \(\left(c_{1} \text { and } c_{2}\right.\) are arbitrary constants), and state the maximum interval over which the solution is valid. $$y(x)=c_{1} e^{-5 x}+c_{2} e^{5 x}, \quad y^{\prime \prime}-25 y=0$$.

Step-by-Step Solution

Answer

The function \(y(x) = c_1e^{-5x} + c_2e^{5x}\) is indeed a solution of the given differential equation \(y'' - 25y = 0\), as substituting and simplifying the derivatives yields an equation of the form \(0 = 0\). The solution is valid on the entire real line, so the maximum interval is (-∞, ∞).

Step 1: Find the first derivative of y(x)

The first step is to find the first derivative, \(y'(x)\): Using the power rule and the chain rule, we have: \[y'(x) = -5c_1e^{-5x} + 5c_2e^{5x}\]

Step 2: Find the second derivative of y(x)

Next, we need to find the second derivative, \(y''(x)\): Applying the power and chain rule again, we get: \[y''(x) = 25c_1e^{-5x} + 25c_2e^{5x}\]

Step 3: Substitute y(x) and y''(x) into the given differential equation

Now, substitute y(x) and y''(x) into the given differential equation, \(y'' - 25y = 0\): \[25c_1e^{-5x} + 25c_2e^{5x} - 25(c_1e^{-5x} + c_2e^{5x}) = 0\]

Step 4: Simplify the expression

Simplify the above expression: \[25c_1e^{-5x} + 25c_2e^{5x} - 25c_1e^{-5x} - 25c_2e^{5x} = 0\] \[0 = 0\] Since the equation is satisfied, the given function \(y(x) = c_1e^{-5x} + c_2e^{5x}\) is indeed a solution of the given differential equation \(y'' - 25y = 0\).

Step 5: Find the maximum interval of the solution

Since the coefficients of the function and its derivatives are continuous for all x, the solution \(y(x) = c_1e^{-5x} + c_2e^{5x}\) is valid for any interval of x, or in other words, on the entire real line (-∞, ∞). Thus, the maximum interval over which the solution is valid is (-∞, ∞).

Key Concepts

Solution Verification

Verifying that a function is a solution to a differential equation involves proving that it satisfies the equation when substituted back in. Consider the function provided:

• The solution function is: \( y(x) = c_1 e^{-5x} + c_2 e^{5x} \), where \( c_1 \) and \( c_2 \) are constants.
• The differential equation to verify against is: \( y'' - 25y = 0 \).

By calculating the derivatives and substituting them back, we check if both sides of the equation match. If they do, our function is indeed a solution. After substituting and simplifying, the expression reduced to \( 0 = 0 \), confirming the function as a valid solution. Understanding this step is crucial because any mistakes in calculation or simplification could lead to incorrect conclusions.

Homogeneous Differential Equation

A homogeneous differential equation is one in which every term can be characterized by its equality in variables and derivatives. In this context, the given equation is:

• \( y'' - 25y = 0 \), which is homogeneous.

Homogeneous equations generally present solutions composed of exponential terms. Here, the characteristic equation derived from \( r^2 - 25 = 0 \), leads to complex roots that manifest as exponential solutions, a hallmark feature of homogeneous equations. Recognizing this type of equation helps in predicting the form a solution might take, often simplifying the solution-finding process significantly for students.

Exponential Function Solution

The presence of exponential functions in the solution, such as \( e^{-5x} \) and \( e^{5x} \), is because of the characteristic equation solutions. For homogeneous linear differential equations with constant coefficients,

• Solutions usually consist of exponential functions.
• The exponents are often roots of the characteristic equation derived from the differential equation.

Exponential functions are important because they define how the solutions to differential equations behave. They can describe a wide range of phenomena in physics and engineering where growth and decay are key characteristics. In our case, the exponential terms in the solution essentially represent these natural dynamic processes.

Maximum Interval of Solution Validity

The interval over which a solution to a differential equation is valid often depends on the points at which the equation or its coefficients become undefined or discontinuous. In the given example:

• The solution \( y(x) = c_1 e^{-5x} + c_2 e^{5x} \) is valid for all real numbers.
• This is because the coefficients, involving constants and exponential functions, remain continuous and well-defined across the entire real line.

The maximum interval for validity in this case is \((-∞, ∞)\). This implies that there is no restriction on \( x \) and the behavior and properties of the solution remain consistent across this interval. Recognizing the validity interval is key in understanding the potential applications and limitations of a solution in real-world scenarios.