Problem 45 - Differential Equations and Linear Algebra

Question

Verify that the given function is a solution to the given differential equation. In these problems, \(c_{1}\) and \(c_{2}\) are arbitrary constants. $$\diamond y(x)=c_{1} e^{2 x}+c_{2} e^{-3 x}, y^{\prime \prime}+y^{\prime}-6 y=0$$.

Step-by-Step Solution

Answer

To verify that the given function \(y(x)=c_{1} e^{2x}+c_{2} e^{-3x}\) is a solution to the given differential equation \(y''(x)+y'(x)-6y(x)=0\), we calculated the first and second derivatives of the function and substituted them into the equation. After simplifying, we found that the equation holds, confirming that \(y(x)=c_{1} e^{2x}+c_{2} e^{-3x}\) is indeed a solution to the given differential equation.

Step 1: Calculate the first derivative of the given function

The given function is \(y(x)=c_{1} e^{2x}+c_{2} e^{-3x}\). To find the first derivative, differentiate each term in the function with respect to \(x\): \[y'(x)=c_{1} \frac{d}{dx} (e^{2x}) + c_{2}\frac{d}{dx} (e^{-3x})\] Using the chain rule, we get \[y'(x)=2c_{1} e^{2x} - 3c_{2} e^{-3x}\]

Step 2: Calculate the second derivative of the given function

Now, let's differentiate the first derivative to obtain the second derivative: \[y''(x)=c_{1} \frac{d^{2}}{dx^{2}} (e^{2x}) + c_{2}\frac{d^{2}}{dx^{2}} (e^{-3x})\] Again, using the chain rule: \[y''(x)=4c_{1} e^{2x} + 9c_{2} e^{-3x}\]

Step 3: Substitute the function and its derivatives into the given equation

We have the equation, \[y''(x)+y'(x)-6y(x)=0\] Substitute the given function and its derivatives: \[(4c_{1} e^{2x} + 9c_{2} e^{-3x})+(2c_{1} e^{2x} - 3c_{2} e^{-3x})-6(c_{1} e^{2x}+c_{2} e^{-3x})=0\]

Step 4: Check if the equation holds

Now, let's simplify the equation: \[(4c_{1} e^{2x} + 9c_{2} e^{-3x})+(2c_{1} e^{2x} - 3c_{2} e^{-3x})-6(c_{1} e^{2x}+c_{2} e^{-3x})\] \[=(4+2-6)c_{1} e^{2x} + (9-3-6)c_{2} e^{-3x}\] \[=0c_{1} e^{2x}+0c_{2} e^{-3x}\] \[=0\] Since the equation holds, we can conclude that the given function \(y(x)=c_{1} e^{2x}+c_{2} e^{-3x}\) is indeed a solution to the given differential equation \(y''(x)+y'(x)-6y(x)=0\).

Key Concepts

Solution Verification

Verifying a solution to a differential equation involves checking if the function satisfies the equation when substituted. In our exercise, the function is given as \(y(x) = c_1 e^{2x} + c_2 e^{-3x}\). The differential equation is \(y'' + y' - 6y = 0\).
To verify, we follow these steps:

• Calculate derivatives of the function as needed by the equation.
• Substitute the function, and its derivatives, back into the differential equation.
• Simplify the equation to see if the left-hand side equals zero.

If it does, the function is indeed a solution. This process is essential when dealing with theoretical and practical applications of differential equations.

Derivatives

Derivatives represent the rate of change of a function. For our function \(y(x) = c_1 e^{2x} + c_2 e^{-3x}\), we seek the first and second derivatives.
The first derivative, using the chain rule, is \(y'(x) = 2c_1 e^{2x} - 3c_2 e^{-3x}\).
The second derivative is obtained in a similar manner: \(y''(x) = 4c_1 e^{2x} + 9c_2 e^{-3x}\).
Derivatives are crucial in forming and solving differential equations, as they provide the needed expressions for substitution back into the equation. Knowing how to accurately compute derivatives is essential for validating potential solutions.

Exponential Functions

Exponential functions are core in mathematics and science. In this case, they form the basis of our given function \(y(x) = c_1 e^{2x} + c_2 e^{-3x}\).
Key properties include:

• The base, \(e\), is Euler's number, approximately equal to 2.71828.
• Exponential functions where the argument is a linear function of \(x\), like \(e^{2x}\) or \(e^{-3x}\), have straightforward derivatives: the same function multiplied by the derivative of the power.

These functions frequently appear in calculus, especially in solving linear differential equations.

Linear Differential Equation

A linear differential equation includes derivatives of a function without any products or powers of the function and its derivatives.
Our equation is \(y'' + y' - 6y = 0\), a classic example of a second-order linear differential equation.Such equations are imperative in modeling real-world systems:

• The linearity allows for a straightforward solution, often accompanied by particular rules and predictable behavior.
• Solving involves finding the complementary or homogeneous solution first, particularly when no nonhomogeneous terms are present, as in our example.

Understanding linear differential equations is key for students, as these equations underpin many applications in engineering, physics, and beyond.