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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? $$ y^{\prime \prime}+2 y^{\prime}-3 y=0 ; y(x)=c_{1} e^{x}+c_{2} e^{-3 x} $$

Short Answer

Expert verified
All solutions are linear combinations of \( e^x \) and \( e^{-3x} \), showing exponential behavior.

Step by step solution

01

Differentiate the function

Start with the given function \( y(x) = c_1 e^x + c_2 e^{-3x} \). Find the first derivative \( y' \) by applying the derivative rules: \( y'(x) = c_1 e^x - 3c_2 e^{-3x} \). Next, find the second derivative \( y'' \): \( y''(x) = c_1 e^x + 9c_2 e^{-3x} \).
02

Substitute into the differential equation

Now that we have \( y' \) and \( y'' \), substitute them into the given differential equation \( y'' + 2y' - 3y = 0 \). This results in: \( (c_1 e^x + 9c_2 e^{-3x}) + 2(c_1 e^x - 3c_2 e^{-3x}) - 3(c_1 e^x + c_2 e^{-3x}) = 0 \).
03

Simplify the expression

Expand and simplify the expression: \( c_1 e^x + 9c_2 e^{-3x} + 2c_1 e^x - 6c_2 e^{-3x} - 3c_1 e^x - 3c_2 e^{-3x} \). This simplifies to: \( (c_1 + 2c_1 - 3c_1)e^x + (9c_2 - 6c_2 - 3c_2)e^{-3x} \). Further simplification gives: \( 0 \). This verifies that the expression satisfies the differential equation.
04

Graph the solutions for several values of \( c_1 \) and \( c_2 \)

Use a graphing utility to plot \( y(x) = c_1 e^x + c_2 e^{-3x} \) for different values of \( c_1 \) and \( c_2 \). For example, try \( c_1 = 1, c_2 = 0 \), \( c_1 = 0, c_2 = 1 \), and \( c_1 = 1, c_2 = 1 \). Note the general behavior of the curves.
05

Analyze the common characteristics of the solutions

Observe that, regardless of the specific values chosen for \( c_1 \) and \( c_2 \), all solutions tend to diverge as \( x \) increases or decreases, depending on the signs of \( c_1 \) and \( c_2 \). However, all solutions are linear combinations of the fundamental solutions \( e^x \) and \( e^{-3x} \).

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Key Concepts

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

Particular Solutions
A particular solution in the context of differential equations is a specific solution derived from the general solution by assigning specific values to the arbitrary constants present. In our given differential equation, the general solution is expressed as: \[ y(x) = c_1 e^x + c_2 e^{-3x} \]where \(c_1\) and \(c_2\) are arbitrary constants. To identify a particular solution, specific values for \(c_1\) and \(c_2\) are chosen, such as \(c_1 = 1\) and \(c_2 = 0\), which results in the solution \(y(x) = e^x\). Each selection of \(c_1\) and \(c_2\) will yield a different curve that satisfies the differential equation.
  • Particular solutions are useful for modeling scenarios with specific initial conditions.
  • The process involves personalization of a general solution to fit particular needs.
Understanding these particular solutions is crucial because they demonstrate how changes in conditions affect the behavior of solutions visually and conceptually.
Graphing Utility
A graphing utility is a powerful tool that helps in visualizing solutions of differential equations. By using a graphing calculator or software like Desmos or GeoGebra, you can instantly see how different values of \(c_1\) and \(c_2\) shape the curve of the solutions.
  • Graphing utilities provide a visual approach that enhances understanding.
  • They allow for dynamic interaction and manipulation of variables to observe results.
  • With graphing utilities, you can explore the behavior of particular solutions over a range.
To use a graphing utility for our differential equation, input the function \[ y(x) = c_1 e^x + c_2 e^{-3x} \]. Try different values for \(c_1\) and \(c_2\), such as \(c_1 = 1, c_2 = 1\) to see the combined effects of these terms. Observing these curves can tell us a lot about the tendencies of solutions, like exponential growth or decay.
Fundamental Solutions
Fundamental solutions are the building blocks of the general solution to a linear homogeneous differential equation. In this example, the differential equation is:\[ y'' + 2y' - 3y = 0 \]The functions \(e^x\) and \(e^{-3x}\) are known as fundamental solutions because they form the basis for constructing all possible solutions to this type of differential equation. Linear combinations of these two solutions, such as \(c_1 e^x + c_2 e^{-3x}\), will satisfy the differential equation for any values of \(c_1\) and \(c_2\).
  • Understanding fundamental solutions is essential for grasping the full scope of differential equations.
  • They illustrate the underlying structure and symmetries of solution spaces.
  • These solutions represent independent behaviors that combined form a wide array of possible solutions.
When analyzing solutions, recognizing their fundamental components helps in simplifying models and predicting long-term behaviors. This understanding aids in fields as varied as physics, engineering, and economics.

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