Question

Verify that the following functions are solutions to the given differential equation.\(y=e^{3 x}-\frac{e^{x}}{2}\) solves \(y^{\prime}=3 y+e^{x}\)

Short answer

The function \(y = e^{3x} - \frac{e^x}{2}\) is a solution to the differential equation \(y' = 3y + e^x\).

Step-by-step solution

Find the Derivative

First, we need to find the derivative of the given function. The function is \( y = e^{3x} - \frac{e^x}{2} \). The derivative is calculated as follows:The derivative of \( e^{3x} \) with respect to \( x \) is \( 3e^{3x} \) using the chain rule. The derivative of \( -\frac{e^x}{2} \) with respect to \( x \) is \( -\frac{1}{2}e^x \) using the constant factor rule.Therefore, the derivative \( y' \) is:\[ y' = 3e^{3x} - \frac{1}{2}e^x \]

Substitute into the Differential Equation

Next, we substitute the function and its derivative into the given differential equation \( y' = 3y + e^x \). We substitute \( y' = 3e^{3x} - \frac{1}{2}e^x \) for \( y' \) and \( y = e^{3x} - \frac{e^x}{2} \):\[ 3e^{3x} - \frac{1}{2}e^x = 3(e^{3x} - \frac{e^x}{2}) + e^x \]

Simplify the Equation

Now, simplify the right side of the equation from Step 2:\[ 3(e^{3x} - \frac{e^x}{2}) + e^x = 3e^{3x} - \frac{3e^x}{2} + e^x \]Combine like terms:\[ 3e^{3x} - \frac{3e^x}{2} + e^x = 3e^{3x} - \frac{3e^x}{2} + \frac{2e^x}{2} \]Further simplify:\[ 3e^{3x} - \frac{1e^x}{2} \]

Verify the Solution

Compare the final simplified expression from the right side of the equation, \( 3e^{3x} - \frac{1}{2}e^x \), with the left side, which is \( y' = 3e^{3x} - \frac{1}{2}e^x \). They are equal, confirming that the given function satisfies the differential equation.

Key concepts

Derivatives

Derivatives are a fundamental concept in calculus, used to measure how a function changes as its input changes. In simpler terms, a derivative tells us the rate of change or the slope of the function at any given point. This concept is vital when solving differential equations, which involve functions and their derivatives.
For example, in the function provided, we have two components: \( e^{3x} \) and \(-\frac{e^x}{2}\). The goal is to find the derivative of the entire function. By exploring each part, we apply rules for differentiation to find how the function changes as \( x \) changes. This calculation will help us verify whether the given function is indeed a solution to the differential equation.
Remember that finding derivatives is the first step when tackling differential equations. It provides a critical link between the original function and the equation we strive to solve.

Chain Rule

The chain rule is a crucial technique in calculus for differentiating composite functions. A composite function is one where one function is nested inside another, such as \( e^{3x} \). The chain rule helps us deal with these situations by allowing us to differentiate the outer function while multiplying it by the derivative of the inner function.
In our exercise, the term \( e^{3x} \) is an example where the chain rule applies. The function \( f(x) = 3x \) multiplied by \( e^x \) presents itself as a candidate for this rule. Here, the outer function is \( e^u \), where \( u = 3x \), and the derivative of \( u \) is simply \( 3 \). Therefore, the derivative is \( 3e^{3x} \).
This rule is instrumental in accurately finding derivatives of nested functions and is a common method used in solving differential equations.

Constant Factor Rule

The constant factor rule simplifies the process of differentiating functions that have constant multipliers. According to this rule, if you have a constant multiplied by a function, you can differentiate the function first and then multiply the derivative by that constant.
In the provided function \(-\frac{e^x}{2}\), there is a constant factor of \(-\frac{1}{2}\). Using the constant factor rule, we first differentiate \(e^x\), which gives us \(e^x\). Then, we multiply by the constant \(-\frac{1}{2}\), resulting in \(-\frac{1}{2}e^x\).
This rule eases the differentiation process, making it straightforward when dealing with multipliers. It's a handy tool for dealing with terms that make up a part of larger differential equations.

Verification of Solutions

Verification of solutions in differential equations is the process of confirming that a proposed solution satisfies the given equation. This is done by substituting the solution and its derivative back into the original differential equation and simplifying the results.
In our exercise, after finding the derivative \( y' = 3e^{3x} - \frac{1}{2}e^x \), we substitute \( y \) and \( y' \) into the equation \( y' = 3y + e^x \).

• First, we substitute \( y = e^{3x} - \frac{e^x}{2} \) into the differential equation.
• Next, we substitute the calculated \( y' \).
• Finally, we simplify and compare both sides of the equation to ensure they match.

If the expression obtained by substituting \( y \) into the differential equation is identical to the expression of \( y' \), we verify the accuracy of the solution, ensuring that the function solves the differential equation. This important step confirms correctness and aids in understanding the relationship between the function and the differential equation.