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Magazine Chapter 31: Problem 34
Show that the given equation is a solution of the given differential equation. $$\frac{d^{3} y}{d x^{3}}+4 \frac{d^{2} y}{d x^{2}}+4 \frac{d y}{d x}=0, \quad y=c_{1}+c_{2} e^{-2 x}+x e^{-2 x}$$
Short Answer
Expert verified
The function satisfies the differential equation because the terms simplify to 0.
Step by step solution
01
Differentiate the given function
The given function is \( y = c_1 + c_2 e^{-2x} + x e^{-2x} \). First, let's find the first derivative \( \frac{dy}{dx} \). \[ \frac{dy}{dx} = \frac{d}{dx}(c_1 + c_2 e^{-2x} + x e^{-2x}) = 0 + c_2(-2)e^{-2x} + e^{-2x} - 2xe^{-2x} = -2c_2 e^{-2x} + e^{-2x} - 2xe^{-2x} \] \[ \frac{dy}{dx} = (-2c_2 + 1 - 2x) e^{-2x} \].
02
Compute the second derivative
Now, differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2 y}{dx^2} \).\[ \frac{d^2 y}{dx^2} = \frac{d}{dx}((-2c_2 + 1 - 2x) e^{-2x}) = (2c_2 - 1 + 2x) (-2) e^{-2x} + (-2)e^{-2x} \] \[ = ((-2c_2 + 1 - 2x) imes (-2) imes e^{-2x} - 2e^{-2x}) \] \[ = ((4c_2 - 2 + 4x) - 2) e^{-2x} \] \[ = (4c_2 + 4x - 4) e^{-2x} \].
03
Compute the third derivative
Differentiate \( \frac{d^2 y}{dx^2} \) to find \( \frac{d^3 y}{dx^3} \).\[ \frac{d^3 y}{dx^3} = \frac{d}{dx}((4c_2 + 4x - 4) e^{-2x}) \] \[ = (-8c_2 - 8x + 8) e^{-2x} + 4 e^{-2x} \] \[ = (-8c_2 - 8x + 8 + 4) e^{-2x} \] \[ = (-8c_2 - 8x + 12) e^{-2x} \].
04
Substitute derivatives into the differential equation
Substitute \( \frac{d^3 y}{dx^3}, \frac{d^2 y}{dx^2}, \frac{dy}{dx} \) into the given differential equation: \[ \frac{d^3 y}{d x^3} + 4 \frac{d^2 y}{d x^2} + 4 \frac{d y}{d x} = 0 \]Substitute the derivatives:\[ (-8c_2 - 8x + 12) e^{-2x} + 4( (4c_2 + 4x - 4) e^{-2x}) + 4((1 - 2x - 2c_2) e^{-2x}) = 0 \].
05
Simplify and verify the equation
Simplify the expression:\[ (-8c_2 - 8x + 12) e^{-2x} + (16c_2 + 16x - 16) e^{-2x} + (4 - 8x - 8c_2) e^{-2x} = 0 \]Combine the terms:\[ (-8c_2 + 16c_2 - 8c_2 - 8x + 16x - 8x + 12 - 16 + 4) e^{-2x} = 0 \]This results in:\[ 0 \cdot e^{-2x} = 0 \], therefore the equation is satisfied.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Third Order Differential Equation
In mathematics, a third order differential equation involves derivatives up to the third order. Essentially, it expresses how a function depends on its derivatives up to three steps.
In the context of our exercise, the given differential equation is:
In the context of our exercise, the given differential equation is:
- \( \frac{d^{3} y}{d x^{3}} + 4 \frac{d^{2} y}{d x^{2}} + 4 \frac{d y}{d x} = 0 \)
Derivative Calculation
Calculating derivatives is crucial for working with differential equations. In our exercise, we need to determine the first, second, and third derivatives of the function \( y = c_1 + c_2 e^{-2x} + x e^{-2x} \).
- The first derivative \( \frac{dy}{dx} \) gives the rate at which \( y \) changes with \( x \), capturing its immediate behavior or slope at any given point.
- For the second derivative \( \frac{d^2 y}{dx^2} \), we see how the rate of change itself changes, offering insights into the curvature or shape of \( y \).
- The third derivative, \( \frac{d^3 y}{dx^3} \), is even more nuanced, revealing the rate of change of the curvature of \( y \).
Exponential Functions
Exponential functions like those found in our exercise, particularly \( e^{-2x} \), are significant due to their rapid growth or decay characteristics.
These functions are defined by the base \( e \), an irrational and transcendental number approximately equal to 2.71828.
These functions are defined by the base \( e \), an irrational and transcendental number approximately equal to 2.71828.
- The function \( e^{-2x} \) specifically describes exponential decay, where the rate of decrease is proportional to its current value.
- Such decay is constant, evidenced by the fact that derivatives of exponential functions are essentially scaled versions of themselves 鈥� no matter how many times you take the derivative.
Equation Verification
Verifying that a proposed solution satisfies a differential equation is vital. It involves substituting derivatives back into the original equation and ensuring all terms simplify correctly.
- In our case, replacing the computed derivatives \( \frac{dy}{dx} \), \( \frac{d^2 y}{dx^2} \), and \( \frac{d^3 y}{dx^3} \) into the equation forms a complex expression.
- Simplifying these terms to check they equate to zero is the final step of verification, confirming our solution's correctness.
