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

Step-by-Step Solution

Answer

The graphs of the solutions all exponentially decay to zero as \( x \rightarrow \\infty \).

Step 1: Differentiate the Given Function

The function given is \( y(x) = c_1 e^{-7x} + c_2 x e^{-7x} \). First, we find the first and second derivatives. Differentiating the first term: \( \frac{d}{dx}[c_1 e^{-7x}] = -7c_1 e^{-7x} \).For the second term, use the product rule \( \frac{d}{dx}[uv] = u'v + uv' \) where \( u = c_2 x \) and \( v = e^{-7x} \):\( u' = c_2, \; v' = -7e^{-7x} \) \( \Rightarrow \frac{d}{dx}[c_2 x e^{-7x}] = c_2 e^{-7x} + c_2 x (-7 e^{-7x}) = c_2 e^{-7x} - 7 c_2 x e^{-7x} \).Combine the derivatives: \( y'(x) = -7c_1 e^{-7x} + c_2 e^{-7x} - 7c_2 x e^{-7x} \).

Step 2: Find the Second Derivative

Now, differentiate \( y'(x) \) again:1. Differentiate \( -7c_1 e^{-7x} \): \( \frac{d}{dx}[-7c_1 e^{-7x}] = 49c_1 e^{-7x} \).2. Differentiate \( c_2 e^{-7x} \): \( \frac{d}{dx}[c_2 e^{-7x}] = -7c_2 e^{-7x} \).3. Differentiate \( -7c_2 x e^{-7x} \) using product rule:\( \frac{d}{dx}[-7c_2 x e^{-7x}] = -7c_2 e^{-7x} + 49c_2 x e^{-7x} \).Combine the terms:\( y''(x) = 49c_1 e^{-7x} - 7c_2 e^{-7x} - 7c_2 e^{-7x} + 49c_2 x e^{-7x} \).Simplify:\( y''(x) = 49c_1 e^{-7x} - 14c_2 e^{-7x} + 49c_2 x e^{-7x} \).

Step 3: Substitute and Verify the Differential Equation

Substitute \( y(x) \), \( y'(x) \), and \( y''(x) \) into the differential equation \( y'' + 14y' + 49y = 0 \):1. Substitute \( y(x) \): \( 49(c_1 e^{-7x} + c_2 x e^{-7x}) \).2. Substitute \( y'(x) \): \( 14(-7c_1 e^{-7x} + c_2 e^{-7x} - 7c_2 x e^{-7x}) \).3. Add \( y''(x) = 49c_1 e^{-7x} - 14c_2 e^{-7x} + 49c_2 x e^{-7x} \).Combine:\[ 49c_1 e^{-7x} - 14c_2 e^{-7x} + 49c_2 x e^{-7x} + 14(-7c_1 e^{-7x} + c_2 e^{-7x} - 7c_2 x e^{-7x}) + 49(c_1 e^{-7x} + c_2 x e^{-7x}) = 0\].After simplification, all terms cancel out, confirming \( y'' + 14y' + 49y = 0 \).

Step 4: Graphing the Particular Solutions

Use a graphing utility to plot the function \( y(x) = c_1 e^{-7x} + c_2 x e^{-7x} \) for different values of \( c_1 \) and \( c_2 \). Choose values such as \( c_1 = 1, c_2 = 0 \); \( c_1 = 0, c_2 = 1 \); \( c_1 = 1, c_2 = 1 \) and observe the behavior of the curves.The common feature of these solutions is that all curves exponentially decay as \( x \rightarrow \infty \), which is due to the presence of the exponential decay factor \( e^{-7x} \).

Key Concepts

Second Order Differential Equations

Second order differential equations involve derivatives up to the second degree. They are important in modeling situations where acceleration or curvature is involved, such as in physics and engineering. These equations generally take the form \( a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = g(x) \), where \(a\), \(b\), and \(c\) are constants, and \(g(x)\) is a function or zero.
Solving these equations often involves methods like characteristic equations, undetermined coefficients, or variation of parameters, depending on whether the equation is homogeneous or non-homogeneous. In our exercise, the equation is homogeneous (no \(g(x)\) term), which simplifies to finding solutions such as exponential or trigonometric functions.

• Homogeneous equations: The right side equals zero, i.e., \(g(x) = 0\).
• Non-homogeneous equations: The right side equals a function of \(x\) i.e., \(g(x) eq 0\).

In the given exercise, solving the homogeneous equation \( y'' + 14y' + 49y = 0 \) involves finding functions whose derivatives satisfy the equation, in this case, yielding solutions involving exponential functions.

Particular Solutions

Particular solutions are specific solutions to a differential equation that satisfy given initial conditions or constraints. While the general solution of a differential equation typically contains arbitrary constants, particular solutions involve specific values that make the equation true for certain conditions.
In this exercise, the general solution \(y(x) = c_1 e^{-7x} + c_2 x e^{-7x}\) is derived from solving the second order differential equation. However, for different values of \(c_1\) and \(c_2\), we obtain particular solutions. These solutions depict the unique behavior of the function under specific circumstances that are defined by initial conditions.

• General solution: Contains arbitrary constants and represents a family of solutions.
• Particular solution: Specific version of the general solution after implementing particular values for the constants.

In practical applications, particular solutions are used when the system's initial condition is given, allowing us to predict future behavior uniquely.

Exponential Decay

Exponential decay refers to a process where quantities reduce at a rate proportional to their current value, often described mathematically by a decaying exponential function like \(e^{-kx}\), where \(k\) is positive. This concept is pivotal in various scientific fields including physics, chemistry, and finance.
In our exercise, the term \(e^{-7x}\) represents exponential decay. It demonstrates how solutions to the differential equation decrease to zero as \(x\) approaches infinity. The exponential decay characteristic is why all graphs derived from varying \(c_1\) and \(c_2\) show a distinct trend downward. This ensures that any initial effects in the system diminish over time.

• Initial high value: The function starts strong but decreases quickly with increasing \(x\).
• As \(x \to \infty\): The value approaches zero, reflecting decay or loss over time.

This feature is significant in modeling natural processes where reduction over time is a key factor, such as radioactive decay, cooling of substances, and depreciation of assets.