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Magazine Chapter 27: Problem 44
Solve the given problems. Show that \(y=e^{-x} \sin x\) satisfies the equation \(\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+2 y=0\)
Short Answer
Expert verified
The function \( y = e^{-x} \sin x \) satisfies the differential equation.
Step by step solution
01
Find the First Derivative
Start with the function given, \( y = e^{-x} \sin x \). To find the first derivative \( \frac{dy}{dx} \), use the product rule: if \( u = e^{-x} \) and \( v = \sin x \), then \( \frac{dy}{dx} = \frac{du}{dx}v + u\frac{dv}{dx} \).Calculate \( \frac{du}{dx} = -e^{-x} \) and \( \frac{dv}{dx} = \cos x \). So, \( \frac{dy}{dx} = (-e^{-x})\sin x + e^{-x} \cos x = e^{-x} (\cos x - \sin x) \).
02
Find the Second Derivative
Now, differentiate the first derivative to find the second derivative. Apply the product rule again to \( \frac{dy}{dx} = e^{-x} (\cos x - \sin x) \). Let \( u = e^{-x} \) and \( v = \cos x - \sin x \).First, find \( \frac{dv}{dx} = -\sin x - \cos x \) and \( \frac{du}{dx} = -e^{-x} \). Then, \( \frac{d^2 y}{d x^2} = (-e^{-x}) (\cos x - \sin x) + e^{-x}(-\sin x - \cos x) = e^{-x}(-2 \cos x) = -2e^{-x} \cos x \).
03
Substitute Derivatives into the Equation
Substitute \( \frac{d^2 y}{dx^2}, \frac{dy}{dx}, \) and \( y \) into the differential equation: \( \frac{d^2 y}{d x^{2}}+2 \frac{d y}{d x}+2 y=0 \).Substitute \( \frac{d^2 y}{dx^2} = -2e^{-x} \cos x \), \( 2 \frac{dy}{dx} = 2e^{-x}(\cos x - \sin x) \), and \( 2y = 2e^{-x} \sin x \) to get:\[-2e^{-x} \cos x + 2e^{-x} (\cos x - \sin x) + 2e^{-x} \sin x = 0.\]
04
Simplify the Expression
Combine like terms in the equation obtained in Step 3. \[-2e^{-x} \cos x + 2e^{-x} \cos x - 2e^{-x} \sin x + 2e^{-x} \sin x = 0.\]Notice that \(-2e^{-x} \cos x + 2e^{-x} \cos x\) and \(-2e^{-x} \sin x + 2e^{-x} \sin x\) both cancel out to yield a sum of 0, confirming the equation is satisfied.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Product Rule
The product rule is a crucial tool in calculus when you need to find the derivative of a product of two functions. Imagine you have two functions, say, \( u \) and \( v \). If these functions are multiplied together, like in the problem where \( y = e^{-x} \sin x \), the product rule helps you find the derivative.Here's how it works:
- Identify the two components, where \( u = e^{-x} \) and \( v = \sin x \).
- Calculate the derivatives of each: \( \frac{du}{dx} = -e^{-x} \) and \( \frac{dv}{dx} = \cos x \).
- Use the formula for the product rule: \( \frac{d}{dx}(uv) = \frac{du}{dx}v + u\frac{dv}{dx} \).
- Apply this to get: \( \frac{dy}{dx} = (-e^{-x})\sin x + e^{-x} \cos x \).
First Derivative
Understanding the first derivative is key to analyzing how a function changes. The first derivative of a function represents the slope of the tangent line at any given point. In simple terms, it's how quickly the function is increasing or decreasing.In our example, with \( y = e^{-x} \sin x \), the first derivative is found using the product rule. After calculation, it simplifies to \( \frac{dy}{dx} = e^{-x} (\cos x - \sin x) \).Here's what we did:
- Applied the product rule to the function.
- Calculated the individual derivatives \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).
- Combined results into the product rule formula.
- Simplified to get the derivative function.
Second Derivative
The second derivative provides deeper insight into the behavior of a function and is essentially the derivative of the first derivative. It helps to assess the function's concavity鈥攚hether the graph curves upwards or downwards.In our function \( y = e^{-x} \sin x \), we find the second derivative of \( \frac{dy}{dx} = e^{-x} (\cos x - \sin x) \) using the product rule again:
- Identify \( u = e^{-x} \) and \( v = \cos x - \sin x \).
- Differentiate again: \( \frac{du}{dx} = -e^{-x} \) and \( \frac{dv}{dx} = -\sin x - \cos x \).
- Apply product rule: \( \frac{d^2 y}{d x^2} = (-e^{-x})(\cos x - \sin x) + e^{-x}(-\sin x - \cos x) = -2e^{-x} \cos x \).
Differential Equation Solution
A differential equation involves functions and their derivatives, capturing a myriad of phenomena around us, from physics to biology. Solving them requires finding a function that satisfies the equation.In this exercise, our task was to show that \( y = e^{-x} \sin x \) satisfies the equation \( \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+2 y=0 \).Here's the process:
- Identify derivatives: \( \frac{d^2 y}{dx^2} = -2e^{-x} \cos x \) and \( \frac{dy}{dx} = e^{-x} (\cos x - \sin x) \).
- Substitute these and the original function into the differential equation.
- Simplify each term to check: \( -2e^{-x} \cos x + 2e^{-x} (\cos x - \sin x) + 2e^{-x} \sin x = 0 \).
- Observe how terms cancel each other out, ensuring the equation balances to zero.
