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In each part, verify that the functions are solutions of the differential equation by substituting the functions into the equation. \(y^{\prime \prime}+y^{\prime}-2 y=0\) (a) \(e^{-2 x}\) and \(e^{x}\) (b) \(c_{1} e^{-2 x}+c_{2} e^{x}\left(c_{1}, c_{2} \text { constants }\right)\)

Short Answer

Expert verified
All given functions are solutions of the differential equation.

Step by step solution

01

Differentiate the functions (a)

For function \(e^{-2x}\):- Compute the first derivative: \(y'(x) = \frac{d}{dx} e^{-2x} = -2e^{-2x}\).- Compute the second derivative: \(y''(x) = \frac{d}{dx}(-2e^{-2x}) = 4e^{-2x}\).For function \(e^{x}\):- Compute the first derivative: \(y'(x) = \frac{d}{dx} e^{x} = e^{x}\).- Compute the second derivative: \(y''(x) = \frac{d}{dx}(e^{x}) = e^{x}\).
02

Substitute into the differential equation (a)

For \(y = e^{-2x}\):Substitute \(y'' = 4e^{-2x}\), \(y' = -2e^{-2x}\), and \(y = e^{-2x}\) into the equation:\[4e^{-2x} + (-2e^{-2x}) - 2(e^{-2x}) = 0\]Which simplifies to:\[4e^{-2x} - 2e^{-2x} - 2e^{-2x} = 0\]So, \(0 = 0\). Thus, \(e^{-2x}\) satisfies the differential equation.For \(y = e^{x}\):Substitute \(y'' = e^{x}\), \(y' = e^{x}\), and \(y = e^{x}\) into the equation:\[e^{x} + e^{x} - 2(e^{x}) = 0\]Which simplifies to:\[e^{x} + e^{x} - 2e^{x} = 0\]So, \(0 = 0\). Thus, \(e^{x}\) satisfies the differential equation.
03

Differentiate the function (b)

For \(y = c_1 e^{-2x} + c_2 e^{x}\):- First derivative: \(y'(x) = c_1 (-2 e^{-2x}) + c_2 (e^{x}) = -2c_1 e^{-2x} + c_2 e^{x}\).- Second derivative: \(y''(x) = c_1 (4 e^{-2x}) + c_2 (e^{x}) = 4c_1 e^{-2x} + c_2 e^{x}\).
04

Substitute into the differential equation (b)

Substitute \(y'' = 4c_1 e^{-2x} + c_2 e^{x}\), \(y' = -2c_1 e^{-2x} + c_2 e^{x}\), and \(y = c_1 e^{-2x} + c_2 e^{x}\) into the equation:\[(4c_1 e^{-2x} + c_2 e^{x}) + (-2c_1 e^{-2x} + c_2 e^{x}) - 2(c_1 e^{-2x} + c_2 e^{x}) = 0\]This simplifies to:\[(4c_1 e^{-2x} - 2c_1 e^{-2x} - 2c_1 e^{-2x}) + (c_2 e^{x} + c_2 e^{x} - 2c_2 e^{x}) = 0\]Which is:\[0 = 0\]This demonstrates that the function \(c_1 e^{-2x} + c_2 e^{x}\) satisfies the differential equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solutions of Differential Equations
A differential equation is a mathematical equation that relates a function to its derivatives. Solving a differential equation means finding the function that satisfies this relationship. In this exercise, we are dealing with a second-order differential equation. This simply means that the highest derivative in the equation is a second derivative, represented by \( y'' \). The equation given is: \( y'' + y' - 2y = 0 \).
This is an example of a linear homogeneous differential equation. To solve it, we need to find functions \( y \) that, when substituted in place of \( y \), \( y' \), and \( y'' \), satisfy the equation completely, resulting in an identity like \( 0 = 0 \).
In this exercise, we are asked to verify if certain functions, such as \( e^{-2x} \) and \( e^{x} \), as well as their linear combination \( c_1 e^{-2x} + c_2 e^{x} \), are valid solutions. The approach involves calculating the derivatives of these functions and substituting them back in the differential equation to check validity.
Verification of Solutions
Verification is the process of ensuring that a given function truly satisfies the differential equation. Even if a function appears correct at first glance, we can only confirm its status as a solution through verification.
Following this exercise's steps, to verify a function, compute its first and second derivatives. For example, with \( y = e^{-2x} \), the first derivative is \( -2e^{-2x} \), and the second is \( 4e^{-2x} \). After substitution back into the differential equation:
  • Substitute \( y'' = 4e^{-2x} \), \( y' = -2e^{-2x} \), and \( y = e^{-2x} \) into \( y'' + y' - 2y = 0 \).
  • Each term combines to result in \( 0 = 0 \).

By reaching this equality, we can confirm that the function is indeed a valid solution. This process confirms that each presented function satisfies the equation according to mathematical rules, including the combination \( c_1 e^{-2x} + c_2 e^{x} \), where constants \( c_1 \) and \( c_2 \) are arbitrary.
Second-order Differential Equation
A second-order differential equation, like the one given in this exercise, is characterized by the presence of a second derivative, \( y'' \). Such equations can describe more complex behaviors in systems, such as oscillations and exponential growth or decay. In general, a second-order differential equation has the form:
  • \( a y'' + b y' + c y = 0 \)

where \( a, b, \) and \( c \) are constants. The simplicity or complexity of the solution can vary based on these constants and the initial conditions, if given. Homogeneous second-order equations are particularly significant because they often appear in physics and engineering, modeling behaviors where there's a balance of forces.
For example, the equation \( y'' + y' - 2y = 0 \), which we worked on, is a second-order linear homogeneous differential equation. Solutions often take the form of exponentials like \( e^{mx} \), where \( m \) is derived from the characteristic equation associated with the differential equation. Learning to solve such equations equips one with skills for analyzing real-world systems.

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