Problem 9 - Calculus Volume 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? $$ [\mathbf{T}] y^{\prime \prime}+14 y^{\prime}+49 y=0 ; \quad y(x)=c_{1} e^{-7 x}+c_{2} x e^{-7 x} $$

Step-by-Step Solution

Answer

The given function is a solution; all solutions decay to zero as \( x \) increases.

Step 1: Understand the Given Differential Equation

The given differential equation is \( y'' + 14y' + 49y = 0 \). This is a second-order linear homogeneous differential equation with constant coefficients.

Step 2: Identify the Given Function

We need to verify if the function \( y(x) = c_1 e^{-7x} + c_2 xe^{-7x} \) is a solution to the differential equation. Pay attention to the terms and parameters involved.

Step 3: Compute the First Derivative

Calculate the first derivative, \( y' \), of the function using the product rule for \( c_2 xe^{-7x} \). This gives:\[ y' = -7c_1 e^{-7x} + c_2 e^{-7x} - 7c_2 xe^{-7x} \].

Step 4: Compute the Second Derivative

Calculate the second derivative, \( y'' \), of the function. Differentiate \( y' = -7c_1 e^{-7x} + c_2 e^{-7x} - 7c_2 xe^{-7x} \) again:\[ y'' = 49c_1 e^{-7x} - 14c_2 e^{-7x} + 49c_2 xe^{-7x} \].

Step 5: Substitute into the Differential Equation

Substitute \( y \), \( y' \), and \( y'' \) into the left side of the original differential equation:\[ y'' + 14y' + 49y = 49c_1 e^{-7x} - 14c_2 e^{-7x} + 49c_2 xe^{-7x} + 14(-7c_1 e^{-7x} + c_2 e^{-7x} - 7c_2 xe^{-7x}) + 49(c_1 e^{-7x} + c_2 xe^{-7x}) \].

Step 6: Simplify the Expression

Simplify the expression from substitution. All terms simplify to zero in this process:\[ (49c_1 e^{-7x} - 98c_1 e^{-7x} + 49c_1 e^{-7x}) + (-14c_2 e^{-7x} + 14c_2 e^{-7x}) + (49c_2 xe^{-7x} - 98c_2 xe^{-7x} + 49c_2 xe^{-7x}) = 0 \].

Step 7: Conclusion on Verification

Since the expression simplifies to zero, the function \( y(x) = c_1 e^{-7x} + c_2 xe^{-7x} \) satisfies the differential equation.

Step 8: Graphing the Solutions

Using a graphing utility, plot \( y(x) = c_1 e^{-7x} + c_2 xe^{-7x} \) for several values of \( c_1 \) and \( c_2 \). The graphs will show that the solutions decay to zero as \( x \) increases.

Key Concepts

Homogeneous Differential Equations

A homogeneous differential equation is a type of differential equation where every term is a function of the dependent variable and its derivatives alone. In simpler terms, there are no terms that are just functions of the independent variable. The given equation \( y'' + 14y' + 49y = 0 \) is a perfect example of this. Here, it’s crucial to recognize that the equation has constant coefficients, which means that the multipliers of each term don’t change. Homogeneous equations often allow us to find special solutions based on certain parameters or constants, leading to a general solution.

To solve these equations, techniques such as the characteristic equation method are typically used. You’ll often find solutions in the form of exponentials, as is the case here, where the general solution is expressed in terms of \( e^{-7x} \). This reflects the damping or decay effect that is common in many physical processes modeled by linear homogeneous differential equations. Understanding these properties can make these equations much more approachable.

Linear Differential Equations

Linear differential equations are characterized by the linearity property: the dependent variable and its derivatives are linear, meaning that there are no cross-products or powers higher than one. This type of equation looks like \( a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + ... + a_1(x)y' + a_0(x)y = g(x) \)where each \( a_i(x) \) is a coefficient that could depend on the independent variable, and \( g(x) \) is a nonhomogeneous term that is zero in our current homogeneous case.

For our specific problem, the equation is not only linear but also homogeneous, making the right-hand side \( g(x) \) equal to zero. Solving these kinds of equations typically involves finding a characteristic polynomial and determining its roots. Each root corresponds to a component of the solution, as seen in \( y(x) = c_1 e^{-7x} + c_2 xe^{-7x} \). Linear differential equations are incredibly important in engineering and physics for modeling systems like electrical circuits or motion under resistance.

Graphing Solutions

Graphing solutions of differential equations provides a visual sense of their behavior over time or space. For example, in the case of \( y(x) = c_1 e^{-7x} + c_2 xe^{-7x} \), the function decreases exponentially as \( x \) grows larger.

Using graphing utilities, such as graphing calculators or software packages, helps us see how different values of \( c_1 \) and \( c_2 \) shape the solution curve. Despite various initial conditions (values of the constants), these solutions often show similar fundamental behavior as seen in decay characteristics. Generally, each curve lies below the x-axis and asymptotically approaches zero for large \( x \).

• This decay reflects the damping effect of exponential decay, common in systems represented by homogeneous differential equations.
• It also provides insight into long-term behavior, crucial in understanding systems stability.

Verification of Solutions

Verification of solutions to differential equations ensures that the proposed solution truly satisfies the equation. This involves substituting the proposed function and its derivatives back into the original differential equation and checking if it holds true.

For our exercise, starting with the function \( y(x) = c_1 e^{-7x} + c_2 xe^{-7x} \), we took its first and second derivatives, then substituted them back into \( y'' + 14y' + 49y = 0 \). Through simplification, every term cancelled out, resulting in zero on both sides of the equation, thus verifying the solution.

• Verification steps are crucial as they provide confidence that the derived solution is correct.
• This process often uncovers potential errors and allows for necessary adjustments.

In this case, by zeroing out the equation upon substitution, it affirmed that our solution satisfies the original homogeneous differential equation for all values of \( c_1 \) and \( c_2 \). Verifying solutions is foundational in solving differential equations accurately and effectively.