A function `y=f(x)` satisfies the differential equation `(dy)/(dx)-y= cosx-sin x` with initial condition that y is bounded when `x _> oo`. The area enclosed by `y=f(x), y=cos x` and the y-axis is

Options

A. (a)`sqrt(2)-1`
B. (b)`sqrt(2)`
C. (c)1
D. (d)`1//sqrt(2)`

Answer

To solve the problem, we need to find the area enclosed by the function \( y = f(x) \), which satisfies the differential equation

\[ \frac{dy}{dx} - y = \cos x - \sin x \]

with the condition that \( y \) is bounded as \( x \to \infty \). We also need to find the area between this function and \( y = \cos x \) along with the y-axis.

### Step 1: Solve the Differential Equation

The given differential equation can be solved using the integrating factor method. The integrating factor \( \mu(x) \) is given by:

\[ \mu(x) = e^{\int -1 \, dx} = e^{-x} \]

Multiplying the entire differential equation by the integrating factor:

\[ e^{-x} \frac{dy}{dx} - e^{-x} y = e^{-x} (\cos x - \sin x) \]

This simplifies to:

\[ \frac{d}{dx}(y e^{-x}) = e^{-x} (\cos x - \sin x) \]

### Step 2: Integrate Both Sides

Integrating both sides with respect to \( x \):

\[ y e^{-x} = \int e^{-x} (\cos x - \sin x) \, dx + C \]

To solve the integral on the right, we can use integration by parts or a known result. The integral of \( e^{-x} \cos x \) and \( e^{-x} \sin x \) can be computed as follows:

\[ \int e^{-x} \cos x \, dx = \frac{e^{-x}}{2} (\sin x + \cos x) \]
\[ \int e^{-x} \sin x \, dx = \frac{e^{-x}}{2} (\sin x - \cos x) \]

Thus, we have:

\[ \int e^{-x} (\cos x - \sin x) \, dx = \frac{e^{-x}}{2} (\sin x + \cos x) - \frac{e^{-x}}{2} (\sin x - \cos x) = e^{-x} \cos x \]

### Step 3: Substitute Back

Substituting back into the equation:

\[ y e^{-x} = e^{-x} \cos x + C \]

Multiplying through by \( e^{x} \):

\[ y = \cos x + C e^{x} \]

### Step 4: Apply the Initial Condition

Since \( y \) must be bounded as \( x \to \infty \), we must have \( C = 0 \) (otherwise, \( C e^{x} \) would grow unbounded). Thus, we find:

\[ y = \cos x \]

### Step 5: Find the Area Enclosed

Now, we need to find the area enclosed by \( y = f(x) = \cos x \), \( y = \cos x \), and the y-axis. Since \( y = f(x) \) is the same as \( y = \cos x \), we need to find the area between \( \cos x \) and itself, which is zero.

However, the area of interest is likely between \( \cos x \) and \( \sin x \) from \( x = 0 \) to \( x = \frac{\pi}{4} \) where they intersect. The area \( A \) can be calculated as:

\[ A = \int_0^{\frac{\pi}{4}} (\cos x - \sin x) \, dx \]

### Step 6: Calculate the Area

Calculating the integral:

\[ A = \int_0^{\frac{\pi}{4}} (\cos x - \sin x) \, dx = \left[ \sin x + \cos x \right]_0^{\frac{\pi}{4}} \]

Evaluating at the bounds:

\[ = \left( \sin \frac{\pi}{4} + \cos \frac{\pi}{4} \right) - \left( \sin 0 + \cos 0 \right) \]
\[ = \left( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \right) - (0 + 1) \]
\[ = \sqrt{2} - 1 \]

### Final Answer

Thus, the area enclosed by \( y = f(x) \), \( y = \cos x \), and the y-axis is:

\[ \boxed{\sqrt{2} - 1} \]