Question

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

Short answer

Yes, the function solves the differential equation.

Step-by-step solution

Understand the Differential Equation

The given differential equation is \( y' = \cos x + y \). This equation states that the derivative of \( y \) with respect to \( x \) should equal the sum of \( \cos x \) and \( y \). Our task is to verify if the function \( y = e^{x} + \frac{\sin x}{2} - \frac{\cos x}{2} \) satisfies this equation.

Compute the Derivative of the Given Function

The given function is \( y = e^{x} + \frac{\sin x}{2} - \frac{\cos x}{2} \). We need to find \( y' \), the derivative of \( y \) with respect to \( x \). The derivative is found by differentiating each term:- The derivative of \( e^{x} \) is \( e^{x} \).- The derivative of \( \frac{\sin x}{2} \) is \( \frac{1}{2}\cos x \).- The derivative of \( -\frac{\cos x}{2} \) is \( \frac{1}{2}\sin x \).Thus, \( y' = e^{x} + \frac{1}{2}\cos x + \frac{1}{2}\sin x \).

Substitute the Function into the Differential Equation

Substitute \( y = e^{x} + \frac{\sin x}{2} - \frac{\cos x}{2} \) into the differential equation \( y' = \cos x + y \):The right-hand side becomes:\[ \cos x + \left( e^{x} + \frac{\sin x}{2} - \frac{\cos x}{2} \right) = e^{x} + \frac{\sin x}{2} + \frac{\cos x}{2} \].

Verify the Equality

Now, compare the expressions we calculated:- Left-hand side (from derivative): \( y' = e^{x} + \frac{1}{2}\cos x + \frac{1}{2}\sin x \).- Right-hand side (from substitution): \( \cos x + y = e^{x} + \frac{1}{2}\sin x + \frac{1}{2}\cos x \).Both sides are equal, hence the given function is indeed a solution to the differential equation.

Key concepts

Function Verification

Verifying whether a particular function is a solution to a differential equation involves substituting the function into the equation and ensuring both sides align mathematically. The differential equation provided here is:\[ y' = \cos x + y \]Our goal is to check if the function:\[ y = e^{x} + \frac{\sin x}{2} - \frac{\cos x}{2} \]satisfies this equation.
To verify this, we perform the following steps:

• Compute the derivative of the function \( y \).
• Substitute both \( y \) and \( y' \) into the differential equation.
• Check if the expressions on both sides of the equation are equal.

By ensuring that both sides match, we confirm that the function is indeed a solution to the differential equation. This process is essential because it shows the function's behavior in relation to the equation.

Derivative Computation

To solve the problem, we need to compute the derivative of the given function.The function:\[ y = e^{x} + \frac{\sin x}{2} - \frac{\cos x}{2} \]consists of three terms, each requiring differentiation:

• The derivative of \( e^{x} \) is \( e^{x} \). Exponential functions like \( e^x \) have derivatives that are directly equal to themselves.
• The derivative of \( \frac{\sin x}{2} \) is \( \frac{1}{2}\cos x \). This uses the rule that the derivative of \( \sin x \) is \( \cos x \), multiplied by the constant \( \frac{1}{2} \).
• Finally, the derivative of \( -\frac{\cos x}{2} \) is \( \frac{1}{2}\sin x \). We apply the rule that the derivative of \( \cos x \) is \(-\sin x \), along with the constant factor.

When combined, the derivative of the whole function \( y \) is:\[ y' = e^{x} + \frac{1}{2}\cos x + \frac{1}{2}\sin x \]Computing derivatives accurately is crucial for verifying functions against differential equations, as it allows us to compare changes in the function with the requirements of the equation.

Trigonometric Functions

Trigonometric functions, like sine and cosine, play a significant role in differential equations. They are periodic, meaning they repeat values in regular intervals, which makes them incredibly versatile in modeling cycles and waves.In the function:\[ y = e^{x} + \frac{\sin x}{2} - \frac{\cos x}{2} \]we see both \( \sin x \) and \( \cos x \). These functions have straightforward rules for differentiation, which are vital for calculating derivatives and verifying our solutions.

• \( \sin x \) differentiates to \( \cos x \).
• \( \cos x \) differentiates to \(-\sin x \).

Understanding these properties allows us to manipulate and use trigonometric functions with ease when dealing with differential equations. In our example, incorporating trigonometric functions with exponential terms helps describe complex behaviors and solutions.