Let `y=f(x)` satisfies the equation `f(x) = (e^(-x)+e^(x))cosx-2x-int_(0)^(x)(x-t)f^(')(t)dt` y satisfies the differential equation

Options

A. `(dy)/(dx)+y=e^(x)(cosx-sinx)-e^(-x)(cosx+sinx)`
B. `(dy)/(dx)-y=e^(x)(cosx-sinx)+e^(-x)(cosx+sinx)`
C. `(dy)/(dx)+y=e^(x)(cosx+sinx)-e^(-x)(cosx-sinx)`
D. `(dy)/(dx) -y=e^(x)(cosx-sinx)+e^(-x)(cosx-sinx)`

Explanation

To solve the problem, we start with the equation given:

\[ f(x) = (e^{-x} + e^{x}) \cos x - 2x - \int_{0}^{x} (x - t) f'(t) dt \]

### Step 1: Rewrite the equation
We can express the integral term in a more manageable form. The equation can be rewritten as:

\[ f(x) = (e^{-x} + e^{x}) \cos x - 2x - \int_{0}^{x} (x f'(t) - t f'(t)) dt \]

### Step 2: Apply integration by parts
We can apply integration by parts to the integral \(\int_{0}^{x} (x - t) f'(t) dt\). Let:

- \(u = x - t\) (first function)
- \(dv = f'(t) dt\) (second function)

Then, we differentiate and integrate:

- \(du = -dt\)
- \(v = f(t)\)

Now, applying integration by parts:

\[ \int u \, dv = uv - \int v \, du \]

This gives us:

\[ \int_{0}^{x} (x - t) f'(t) dt = (x-t)f(t) \bigg|_{0}^{x} - \int_{0}^{x} f(t) (-dt) \]

### Step 3: Evaluate the boundary terms
Evaluating the boundary terms:

\[ (x-x)f(x) - (x-0)f(0) = 0 - x f(0) = -x f(0) \]

Thus, we have:

\[ \int_{0}^{x} (x - t) f'(t) dt = -x f(0) + \int_{0}^{x} f(t) dt \]

### Step 4: Substitute back into the equation
Substituting this back into our equation for \(f(x)\):

\[ f(x) = (e^{-x} + e^{x}) \cos x - 2x + x f(0) - \int_{0}^{x} f(t) dt \]

### Step 5: Differentiate both sides
Now, we differentiate both sides with respect to \(x\):

\[ f'(x) = \frac{d}{dx}[(e^{-x} + e^{x}) \cos x] - 2 + f(0) + x f'(0) - f(x) \]

Using the product rule on the right side:

\[ f'(x) = \left( -e^{-x} \cos x + e^{x} \cos x - (e^{-x} + e^{x}) \sin x \right) - 2 + f(0) + x f'(0) - f(x) \]

### Step 6: Rearranging the equation
Rearranging gives us a differential equation involving \(f(x)\) and \(f'(x)\):

\[ f'(x) + f(x) = -\left( e^{-x} \cos x + e^{x} \cos x - (e^{-x} + e^{x}) \sin x - 2 + f(0) + x f'(0) \right) \]

### Step 7: Solve for \(f(0)\)
To find \(f(0)\), we substitute \(x = 0\) in the original equation:

\[ f(0) = (e^{0} + e^{0}) \cos(0) - 2(0) - 0 = 2 \]

### Conclusion
The function \(f(x)\) satisfies the differential equation derived above. The final form of the differential equation is:

\[ f'(x) + f(x) = -\left( e^{-x} \cos x + e^{x} \cos x - (e^{-x} + e^{x}) \sin x - 2 + 2 + 0 \right) \]