Problem 10 - Calculus 3

Question

For each of the following problems, verify that the given function is a solution to the differential equation. Use a graphing utility to graph the particular solutions for several values of \(c_{1}\) and \(c_{2} .\) What do the solutions have in common? $$ [\mathrm{T}] 6 y^{\prime \prime}-49 y^{\prime}+8 y=0 ; \quad y(x)=c_{1} e^{x / 6}+c_{2} e^{8 x} $$

Step-by-Step Solution

Answer

The function \( y(x) = c_1 e^{x/6} + c_2 e^{8x} \) is a solution; common traits in graphs are exponential growth patterns.

Step 1: Understand the Problem

We are provided with a differential equation \( 6 y'' - 49 y' + 8 y = 0 \) and a potential solution \( y(x) = c_1 e^{x/6} + c_2 e^{8x} \). We need to verify if the given function is indeed a solution to the differential equation for arbitrary constants \( c_1 \) and \( c_2 \).

Step 2: Find the First Derivative

Compute the first derivative of \( y(x) = c_1 e^{x/6} + c_2 e^{8x} \). The derivative is:\[y'(x) = \frac{c_1}{6} e^{x/6} + 8c_2 e^{8x}\]

Step 3: Find the Second Derivative

Find the second derivative of the function. Differentiating \( y'(x) \) gives:\[y''(x) = \frac{c_1}{36} e^{x/6} + 64c_2 e^{8x}\]

Step 4: Substitute into the Differential Equation

Substitute \( y(x) \), \( y'(x) \), and \( y''(x) \) back into the original differential equation. That is,\[ 6\left(\frac{c_1}{36} e^{x/6} + 64c_2 e^{8x}\right) - 49\left(\frac{c_1}{6} e^{x/6} + 8c_2 e^{8x}\right) + 8\left( c_1 e^{x/6} + c_2 e^{8x} \right) = 0 \]Simplify this expression to verify it equals zero.

Step 5: Simplify the Equation

Simplify each term in the substituted equation:- The first term: \( \frac{c_1}{6} e^{x/6} + 384c_2 e^{8x} \)- The second term: \( -\frac{49c_1}{6} e^{x/6} - 392c_2 e^{8x} \)- The third term: \( 8c_1 e^{x/6} + 8c_2 e^{8x} \)Combining these gives:\[ \left( \frac{c_1}{6} - \frac{49c_1}{6} + 8c_1 \right) e^{x/6} + (384c_2 - 392c_2 + 8c_2 ) e^{8x} = 0 \]

Step 6: Verify the Equation Equals Zero

Combining the coefficients:- For \( e^{x/6} \): \( \frac{c_1}{6} - \frac{49c_1}{6} + 8c_1 = 0 \)- For \( e^{8x} \): \( 384c_2 - 392c_2 + 8c_2 = 0 \)Both expressions equal zero, showing that \( y(x) = c_1 e^{x/6} + c_2 e^{8x} \) is a solution.

Step 7: Graphing and Commonality

Use a graphing utility to plot \( y(x) = c_1 e^{x/6} + c_2 e^{8x} \) for different values of \( c_1 \) and \( c_2 \). Notice that for each solution, the behavior and the shape of the curve remain similar, generally showing exponential growth, with transformations depending on \( c_1 \) and \( c_2 \). All graphs will show non-linear exponential behavior and share asymptotic trends.

Key Concepts

Solution Verification

Ensuring a function is a solution to a differential equation is called solution verification. In this context, the process began by understanding the problem at hand. We had a second-order differential equation given by \( 6 y'' - 49 y' + 8 y = 0 \). Our task was to verify that the function \( y(x) = c_1 e^{x/6} + c_2 e^{8x} \) is indeed a solution.The key steps involved in solution verification include:

• Calculating the first and second derivatives of the proposed solution \( y(x) \): This involved taking derivative functions like \( y'(x) = \frac{c_1}{6} e^{x/6} + 8c_2 e^{8x} \) and \( y''(x) = \frac{c_1}{36} e^{x/6} + 64c_2 e^{8x} \).
• Substituting these derivatives, along with the original function, back into the differential equation.
• Ensuring that, after simplification, all terms equate to zero.

This confirms that the given function satisfies the equation for any constants \( c_1 \) and \( c_2 \). Thus, this verification ensures that the function can indeed be a solution to the differential equation.

Exponential Functions

Exponential functions like \( e^{x/6} \) and \( e^{8x} \) are crucial in solving differential equations. They have unique properties that make them suitable for describing diverse phenomena in mathematics and physics.Key characteristics of exponential functions include:

• Continuous Growth or Decay: These functions show continuous change over time, either increasing (growth) or decreasing (decay), depending on the exponent.
• Derivatives Proportional to the Original Function: Notably, the derivative of an exponential function is proportional to the function itself, which makes them particularly useful in solving differential equations.
• Asymptotic Behavior: Exponential functions can approach, but never touch, the x-axis or y-axis, depending on their orientation.

The properties of these functions are leveraged in constructing solutions to differential equations, where they help model dynamic systems and growth processes effectively.

Graphing Solutions

Graphing solutions of differential equations provides a visual representation of how these equations behave under different conditions.To graph the solutions of our differential equation \( y(x) = c_1 e^{x/6} + c_2 e^{8x} \):

• Choose different values of the arbitrary constants \( c_1 \) and \( c_2 \). These constants help modify the path of the graph.
• Utilize graphing utilities or software to plot these expressions, observing their behavior and shape.
• The resulting graphs will typically manifest as curved lines, each showing exponential growth based on the values of \( c_1 \) and \( c_2 \).

Each plotted graph, regardless of the constants chosen, will share common shapes and trends, such as exponential increases or decreases. This visual exercise provides insight into the pattern and general behavior of the solutions. It helps to ensure understanding beyond algebraic manipulation.

Second-Order Differential Equations

Second-order differential equations, like the one given \( 6 y'' - 49 y' + 8 y = 0 \), involve the second derivative of a function. These equations model various physical phenomena, including harmonic oscillators and other systems in engineering and physics.Main features of second-order differential equations include:

• Dependence on Second Derivatives: The second derivative \( y'' \) is a critical aspect that influences the behavior and solutions of the equations.
• Characteristic Equation: Solving involves deriving a characteristic equation whose roots determine the solution form, either as exponential, sine, or cosine functions.
• Homogeneous versus Non-Homogeneous: Equations can be either homogeneous (as in our case, where all terms relate to the function or its derivatives) or non-homogeneous (with additional terms not depending solely on the function).

These equations often require techniques like trial solutions involving exponential or trigonometric functions to find their general solutions. Understanding this type of equation helps in evaluating complex systems and predicting their future states.