Question

Verify that the following functions are solutions to the given differential equation.\(y=2 e^{x}-x-1\) solves \(y^{\prime}=y+x\)

Short answer

The function is a solution to the differential equation.

Step-by-step solution

Identify Given Functions and Equation

We are given the function \( y = 2e^x - x - 1 \) and the differential equation \( y' = y + x \). The task is to verify if the function is a solution to the differential equation.

Differentiate the Given Function

Differentiate the function \( y = 2e^x - x - 1 \) with respect to \( x \). This will give us \( y' = \frac{d}{dx}(2e^x - x - 1) = 2e^x - 1 \).

Substitute into the Differential Equation

Substitute \( y = 2e^x - x - 1 \) and \( y' = 2e^x - 1 \) into the differential equation \( y' = y + x \). So, the left hand side becomes: \( 2e^x - 1 \).

Substitute the Function into the Expression \( y + x \)

Substitute \( y = 2e^x - x - 1 \) into \( y + x \) to get \( y + x = (2e^x - x - 1) + x = 2e^x - 1 \).

Check the Equality

Compare the expressions from Steps 3 and 4. We find that both sides simplify to \( 2e^x - 1 \), which are equal. Thus, \( y = 2e^x - x - 1 \) satisfies the differential equation \( y' = y + x \).

Key concepts

Solution Verification

When we are faced with verifying a solution to a differential equation, it's essentially about checking if our given function satisfies the equation. For this task, it's necessary to first differentiate the given function and see how it relates to the required differential equation. Here's a simple way to tackle this:

• Start by differentiating your function with respect to its variable.
• Then substitute both the original function and its derivative into the differential equation.
• Check if both sides of the equation match. If they do, you have verified that the function is indeed a solution.

In our exercise, the function \(y = 2e^x - x - 1\) has been differentiated to get \(y' = 2e^x - 1\). By inserting these values into the differential equation \(y' = y + x\), both sides computed to \(2e^x - 1\), clearly showing that the function is a valid solution.

Exponential Functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. A common example is \( e^x \), where \( e \) is Euler's number, approximately 2.718. These functions have unique characteristics:

• They grow rapidly and have an ever-increasing derivative.
• The derivative of \( e^x \) is particularly convenient as it remains \( e^x \), making calculations straightforward.
• Exponential functions often model real-world phenomena such as population growth, radioactive decay, and more.

In the context of our exercise, the function \(2e^x\) within \(y = 2e^x - x - 1\) plays a pivotal role in both the differentiation process and in meeting the conditions of the differential equation. This property of maintaining its form upon differentiation simplifies verification tasks.

Differentiation

Differentiation is the mathematical process of finding the derivative of a function, which represents the function's rate of change. This topic is crucially involved in our exercise when determining \(y'\) from the original function \(y = 2e^x - x - 1\). Here's why it matters:

• To differentiate, apply rules such as the power rule, product rule, and chain rule applicable to the expressions involved.
• The derivative of \(e^x\) is simple: it’s \(e^x\), allowing us to process equations efficiently.
• The derivative of \( -x \) is \(-1\), while the derivative of a constant (like \(-1\) here) is 0.

So, the differentiation outcome \(y' = 2e^x - 1\) elegantly sumarizes these rules and the process, which then fits seamlessly into the differential equation. This underlines the power of differentiation in solving and verifying solutions for differential equations.