Problem 11 - Calculus Early Transcendentals

Question

In each part, verify that the functions are solutions of the differential equation by substituting the functions into the equation. $$ \begin{array}{l}{y^{\prime \prime}-4 y^{\prime}+4 y=0} \\ {\text { (a) } e^{2 x} \text { and } x e^{2 x}} \\ {\text { (b) } c_{1} e^{2 x}+c_{2} x e^{2 x}\left(c_{1}, c_{2} \text { constants }\right)}\end{array} $$

Step-by-Step Solution

Answer

Both given functions in parts (a) and (b) are solutions to the differential equation.

Step 1: Identify the Differential Equation

The given differential equation is \( y'' - 4y' + 4y = 0 \). Our task is to verify whether certain proposed functions are solutions of this equation.

Step 2: Verify Solution for Part (a) - Function \( e^{2x} \)

Let's first substitute the function \( y = e^{2x} \) into the differential equation. - Compute the first derivative: \( y' = \frac{d}{dx}(e^{2x}) = 2e^{2x} \) - Compute the second derivative: \( y'' = \frac{d^2}{dx^2}(e^{2x}) = 4e^{2x} \) - Substitute these into the differential equation:\[ y'' - 4y' + 4y = 4e^{2x} - 4(2e^{2x}) + 4e^{2x} = 0 \] This confirms that \( y = e^{2x} \) is a solution.

Step 3: Verify Solution for Part (a) - Function \( xe^{2x} \)

Now substitute \( y = xe^{2x} \) into the differential equation. - First derivative using product rule: \( y' = \frac{d}{dx}(xe^{2x}) = e^{2x} + 2xe^{2x} \)- Second derivative: \( y'' = \frac{d^2}{dx^2}(xe^{2x}) = 2e^{2x} + 2(e^{2x} + 2xe^{2x}) = 4xe^{2x} + 4e^{2x} \)- Substitute into the differential equation:\[ y'' - 4y' + 4y = (4xe^{2x} + 4e^{2x}) - 4(e^{2x} + 2xe^{2x}) + 4xe^{2x} = 0 \]This confirms that \( y = xe^{2x} \) is a solution.

Step 4: Verify Solution for Part (b) - Function \( c_1 e^{2x} + c_2 xe^{2x} \)

Substitute the general solution \( y = c_1 e^{2x} + c_2 xe^{2x} \) into the differential equation for constants \( c_1 \) and \( c_2 \).- First derivative: \( y' = c_1(2e^{2x}) + c_2(e^{2x} + 2xe^{2x}) = 2c_1 e^{2x} + c_2 e^{2x} + 2c_2 xe^{2x} \)- Second derivative: \( y'' = c_1(4e^{2x}) + c_2(2e^{2x} + 2(e^{2x} + 2xe^{2x})) = 4c_1 e^{2x} + 2c_2 e^{2x} + 4c_2 xe^{2x} \)- Substitute into the differential equation:\[ y'' - 4y' + 4y = (4c_1 e^{2x} + 2c_2 e^{2x} + 4c_2 xe^{2x}) - 4(2c_1 e^{2x} + c_2 e^{2x} + 2c_2 xe^{2x}) + 4(c_1 e^{2x} + c_2 xe^{2x}) = 0 \]This confirms that \( y = c_1 e^{2x} + c_2 xe^{2x} \) is a solution.

Key Concepts

Second-Order Differential Equation

A differential equation is an equation involving a function and its derivatives. The order of a differential equation is determined by the highest derivative present. In a second-order differential equation, like the one in our exercise, we involve up to the second derivative of the function.

The given differential equation is: \[ y'' - 4y' + 4y = 0 \]Here, \( y'' \) is the second derivative, \( y' \) is the first derivative, and \( y \) is the original function. Understanding this hierarchy helps us tackle the problem by checking how each piece contributes to the overall equation.

Such equations often appear in physical systems, like oscillations or mechanical vibrations, where forces depend on velocity and acceleration. Recognizing the structure of these equations can help simplify complex real-world tasks.

Solution Verification

In the context of differential equations, verifying solutions involves plugging proposed solutions back into the original equation. Our job is to ensure the equation holds true with those substitutions.

• Calculate derivatives as needed: This often involves finding the first and second derivatives. For example, if \( y = e^{2x} \), the derivatives would be \( y' = 2e^{2x} \) and \( y'' = 4e^{2x} \).
• Substitute these derivatives into the differential equation: Replace all \( y \,\), \( y' \,\), and \( y'' \) in the original equation with their corresponding expressions.
• Simplify: Combine like terms and confirm if the equation equals zero, ensuring the function is indeed a solution.

These steps are crucial for verifying whether given functions are the correct solutions. Successfully doing so confirms that the functions achieve what the differential equation describes.

Homogeneous Equation

A homogeneous differential equation is characterized by each term being dependent on the function or its derivatives, with no external function (or forcing term) added. Mathematically, this means the equation equals zero.

Consider the equation:\[ y'' - 4y' + 4y = 0 \]Each term involves either the function \( y \) or its derivatives \( y' \) or \( y'' \), and there's nothing added to it beyond these homogeneously structured terms.

Homogeneous equations often have solutions that can be expressed in terms of exponential functions. Why? Because exponential functions have the unique property that they remain proportional to their derivatives. This quality makes them naturally suited to this type of differential equation. Understanding this helps recognize potential solutions intuitively, speeding up the solution process.