Problem 47 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, explain why or give an example to show why it is false. The function \(f(x)=2 e^{x}-\frac{1}{2}(\cos x+\sin x)\) is a solution of the differential equation \(y^{\prime}-y=\sin x\).

Step-by-Step Solution

Answer

The statement is False. After substituting \(f(x) = 2e^x - \frac{1}{2}(\cos x + \sin x)\) and its first derivative \(f'(x) = 2e^x - \frac{1}{2}\sin x + \frac{1}{2}\cos x\) into the differential equation \(y' - y = \sin x\), the resulting equation is \(\cos x = \sin x\), which is not equivalent to the original equation. Therefore, the function is not a solution to the given differential equation.

Step 1: Find the first derivative of f(x)

We need to find the first derivative of the function: $$ f(x) = 2e^x - \frac{1}{2}(\cos x+\sin x) $$ Deriving both terms separately, 1. Derivative of \(2e^x\): Using the chain rule, we have $$ \frac{d}{dx}(2e^x) = 2 \cdot e^x $$ 2. Derivative of \(\frac{1}{2}(\cos x + \sin x)\): Using the chain rule, we have $$ \frac{d}{dx}\left(\frac{1}{2}(\cos x + \sin x)\right) = -\frac{1}{2}\sin x+\frac{1}{2}\cos x $$ Adding these results together, we get the first derivative of \(f(x)\): $$ f'(x) = 2e^x - \frac{1}{2}\sin x + \frac{1}{2}\cos x $$

Step 2: Plug f(x) and f'(x) into the differential equation

Now, we plug in \(f(x)\) and \(f'(x)\) into the given differential equation \(y' - y = \sin x\): $$ (2e^x - \frac{1}{2}\sin x + \frac{1}{2}\cos x) - (2e^x - \frac{1}{2}(\cos x + \sin x)) = \sin x $$

Step 3: Simplify the equation

We now simplify the equation and see if it holds true: $$ 2e^x - \frac{1}{2}\sin x + \frac{1}{2}\cos x - 2e^x + \frac{1}{2}\sin x + \frac{1}{2}\cos x = \sin x $$ The terms \(2e^x\) and \(-2e^x\) cancel out, and we are left with: $$ \cos x = \sin x $$

Step 4: Conclusion

Since the differential equation simplifies to \(\cos x = \sin x\), which is not equivalent to the original equation \(y'-y=\sin x\), the statement is False. The function \(f(x) = 2e^x - \frac{1}{2}(\cos x + \sin x)\) is not a solution to the differential equation \(y' - y = \sin x\).