Solutions Of The Linear Differential Equation Y'' - Y = 0

In the realm of differential equations, identifying solutions is a fundamental task. This article delves into the process of determining which functions satisfy a given linear differential equation. Specifically, we will focus on the equation y'' - y = 0 and explore whether the functions e^x, sin x, cos x, and sin x - cos x are solutions. This exploration will not only solidify your understanding of differential equations but also equip you with the skills to tackle similar problems.

Understanding Linear Differential Equations

Before we embark on our solution-finding journey, let's first establish a firm grasp of what linear differential equations are. At its core, a differential equation is an equation that involves an unknown function and its derivatives. A linear differential equation is a special type of differential equation where the unknown function and its derivatives appear linearly. This means that they are not raised to any powers, multiplied by each other, or involved in any nonlinear functions.

Our focal equation, y'' - y = 0, perfectly exemplifies a linear differential equation. Here, y'' represents the second derivative of the unknown function y, and the equation states that the second derivative of y minus y itself equals zero. This seemingly simple equation holds a wealth of mathematical significance and finds applications in diverse fields, ranging from physics to engineering.

To solve a linear differential equation, we seek functions that, when substituted into the equation, make the equation true. These functions are the solutions of the differential equation. Finding these solutions often involves employing various techniques, such as characteristic equations, undetermined coefficients, or variation of parameters. However, in some cases, we can directly verify whether a given function is a solution by substituting it into the equation and checking if it satisfies the equality.

Verifying Solutions: A Step-by-Step Approach

Our primary objective is to determine which of the given functions—e^x, sin x, cos x, and sin x - cos x—are solutions to the differential equation y'' - y = 0. To achieve this, we will adopt a systematic approach that involves the following steps:

1. Calculate the Second Derivative: For each candidate function, we need to compute its second derivative. This involves differentiating the function twice with respect to x.
2. Substitute into the Differential Equation: Once we have the second derivative, we substitute both the original function and its second derivative into the differential equation y'' - y = 0.
3. Check for Equality: After substitution, we simplify the resulting expression and check if it equals zero. If it does, then the function is a solution to the differential equation; otherwise, it is not.

Let's now apply this process to each of the candidate functions.

1. Examining e^x as a Potential Solution

Our first candidate is the exponential function e^x. To verify if it's a solution, we need to compute its second derivative. The first derivative of e^x is simply e^x, and the second derivative is also e^x. Now, we substitute y = e^x and y'' = e^x into the differential equation:

e^x - e^x = 0

The equation holds true, indicating that e^x is indeed a solution to the differential equation y'' - y = 0. This result is not surprising, as exponential functions are known to play a crucial role in solving many linear differential equations.

2. Investigating sin x as a Potential Solution

Next, we consider the sine function, sin x. The first derivative of sin x is cos x, and the second derivative is -sin x. Substituting y = sin x and y'' = -sin x into the differential equation, we get:

-sin x - sin x = -2sin x

This expression does not equal zero for all values of x. Therefore, sin x is not a solution to the differential equation y'' - y = 0. This outcome highlights the importance of careful verification, as not all trigonometric functions will satisfy this particular differential equation.

3. Analyzing cos x as a Potential Solution

Now, let's turn our attention to the cosine function, cos x. The first derivative of cos x is -sin x, and the second derivative is -cos x. Substituting y = cos x and y'' = -cos x into the differential equation, we obtain:

-cos x - cos x = -2cos x

Similar to the case of sin x, this expression does not equal zero for all values of x. Consequently, cos x is not a solution to the differential equation y'' - y = 0. This finding reinforces the notion that the specific form of the function significantly impacts its ability to satisfy a given differential equation.

4. Evaluating sin x - cos x as a Potential Solution

Finally, we examine the function sin x - cos x. To find its second derivative, we first differentiate it to get cos x + sin x. Differentiating again, we obtain -sin x + cos x. Substituting y = sin x - cos x and y'' = -sin x + cos x into the differential equation, we have:

(-sin x + cos x) - (sin x - cos x) = -2sin x + 2cos x

This expression, once again, does not equal zero for all values of x. Thus, sin x - cos x is not a solution to the differential equation y'' - y = 0. This result underscores the fact that linear combinations of functions that are not individually solutions may also not be solutions themselves.

Conclusion: Identifying the Solutions

Through our meticulous analysis, we have determined that only e^x is a solution to the linear differential equation y'' - y = 0 among the given options. The functions sin x, cos x, and sin x - cos x do not satisfy the equation. This exercise demonstrates the importance of verifying potential solutions by substituting them into the differential equation and checking for equality. This process not only confirms whether a function is a solution but also deepens our understanding of the equation's behavior.

The differential equation y'' - y = 0 is a classic example of a second-order linear homogeneous differential equation with constant coefficients. Such equations have wide-ranging applications in various scientific and engineering disciplines. The general solution to this particular equation can be expressed as a linear combination of e^x and e^-x, highlighting the fundamental role of exponential functions in solving these types of equations.

By mastering the techniques presented in this article, you will be well-equipped to tackle a wide array of differential equation problems. Remember, the key is to understand the underlying concepts, apply the appropriate methods, and meticulously verify your solutions. Happy problem-solving!

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