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 is a solution to the differential equation.

Step-by-step solution

Find the Derivative

First, we need to find the derivative of the given function \(y = e^{3x} - \frac{e^x}{2}\). We'll use the basic derivative rules: the derivative of \(e^{ax}\) with respect to \(x\) is \(a e^{ax}\). Therefore, the derivative \(y'\) is the result of \(d/dx(e^{3x}) = 3e^{3x}\) and \(d/dx(-\frac{e^x}{2}) = -\frac{1}{2}e^x\). So, the derivative \(y'\) is:\[ y' = 3e^{3x} - \frac{1}{2}e^x \]

Substitute into the Differential Equation

Next, substitute \(y\) and \(y'\) into the differential equation \(y' = 3y + e^x\). Start by calculating the right-hand side of the equation:\[ 3y + e^x = 3(e^{3x} - \frac{e^x}{2}) + e^x \]This simplifies to:\[ 3e^{3x} - \frac{3}{2}e^x + e^x \]

Simplify the Equation

Now simplify the expression \(3e^{3x} - \frac{3}{2}e^x + e^x\). Combine the \(e^x\) terms:\[ -\frac{3}{2}e^x + e^x = -\frac{3}{2}e^x + \frac{2}{2}e^x = -\frac{1}{2}e^x \]Thus, the right-hand side becomes:\[ 3e^{3x} - \frac{1}{2}e^x \]

Verify the Solution

Compare the expression for \(y' = 3e^{3x} - \frac{1}{2}e^x\) we found in step 1 with the expression for the right-hand side of the differential equation \(3e^{3x} - \frac{1}{2}e^x\). They are identical, confirming that the function \(y = e^{3x} - \frac{e^x}{2}\) is indeed a solution to the differential equation \(y' = 3y + e^x\).

Key concepts

Understanding Derivative Rules

Derivatives are a core concept in calculus, and knowing the rules of differentiation is crucial to solving differential equations. For exponential functions like \( e^{ax} \), the derivative is straightforward. The derivative of \( e^{ax} \) with respect to \( x \), where \( a \) is a constant, is \( ae^{ax} \). This means you multiply the exponential function by the exponent's constant.

Here are some helpful rules to remember when working with derivatives:

• The power rule states that the derivative of \( x^n \) is \( nx^{n-1} \).
• The constant multiple rule lets you pull constants out of the derivative: \( d/dx (cf(x)) = c(d/dx (f(x))) \).
• The addition rule allows you to differentiate sums term by term: \( d/dx (u + v) = du/dx + dv/dx \).

So, when finding the derivative \( y' \) for \( y = e^{3x} - \frac{e^x}{2} \), we apply these rules. Manage each part separately: the derivative of \( e^{3x} \) is \( 3e^{3x} \), and for \( \frac{e^x}{2} \), it is \( -\frac{1}{2}e^x \). Combining these results gives us \( y' = 3e^{3x} - \frac{1}{2}e^x \).

Familiarizing yourself with these rules will ease your understanding of calculus, as they form the foundation of derivative computations.

Solution Verification Process

Verifying a solution to a differential equation involves substituting the proposed solution back into the equation to check its validity. This process ensures that the function you have satisfies the given differential equation.

To check if \( y = e^{3x} - \frac{e^x}{2} \) is a solution to \( y' = 3y + e^x \), proceed as follows:

• First, substitute the function \( y \) into the differential equation. In our example, compute \( 3y + e^x \), and simplify.
• Next, simplify the expression by combining like terms. So, \( 3(e^{3x} - \frac{e^x}{2}) + e^x \) simplifies to \( 3e^{3x} - \frac{3}{2}e^x + e^x \), which eventually reduces to \( 3e^{3x} - \frac{1}{2}e^x \).
• Finally, check if this simplified expression is the same as \( y' \), which is \( 3e^{3x} - \frac{1}{2}e^x \).

If both sides match completely, the function satisfies the differential equation, confirming its correctness. Use this method whenever checking solutions to differential equations.

Exploring Exponential Functions

Exponential functions are prevalent in calculus and differential equations. They usually take the form \( e^{ax} \), where \( e \) is the base of the natural logarithm and \( a \) is a constant. These functions have unique properties that make them important in modeling real-world phenomena such as population growth and radioactive decay.

Let's consider their key characteristics:

• An exponential function grows (or decays) at a rate proportional to its current value. This property is why you'll often see exponential functions in differential equations, which model such phenomena.
• The derivative of an exponential function \( e^{ax} \) is another exponential function, \( ae^{ax} \), meaning its rate of growth is itself an exponential function.
• Exponential functions never touch the x-axis; they asymptotically approach it as \( x \to -\infty \) or \( x \to \infty \), depending on the function's sign and the coefficient \( a \).

Understanding these concepts is crucial, as exponential functions often appear as solutions to differential equations due to their growth properties. Recognizing these functions and their derivatives can dramatically simplify solving differential equations.