The solution of the differential equation, `x^(2)dy/dxcos""(1)/(x)-ysin""(1)/(x)=-1,` where `yto-1asxtooo` is

Options

A. `y=sin""(1)/(x)-cos""(1)/(x)`
B. `y=(x+1)/(xsin""(1)/(x))`
C. `y=sin""(1)/(x)+cos""(1)/(x)`
D. `y=(x+1)/(xcos""(1)/(x))`

Answer

To solve the given differential equation

\[ x^2 \frac{dy}{dx} \cos\left(\frac{1}{x}\right) - y \sin\left(\frac{1}{x}\right) = -1 \]

where \( y \to -1 \) as \( x \to \infty \), we can follow these steps:

### Step 1: Rewrite the Differential Equation
We start by rewriting the equation in a more standard form. We can divide the entire equation by \( x^2 \cos\left(\frac{1}{x}\right) \):

\[ \frac{dy}{dx} - \frac{y \sin\left(\frac{1}{x}\right)}{x^2 \cos\left(\frac{1}{x}\right)} = -\frac{1}{x^2 \cos\left(\frac{1}{x}\right)} \]

### Step 2: Identify \( p(x) \) and \( q(x) \)
From the rewritten equation, we can identify:

\[ p(x) = -\frac{\sin\left(\frac{1}{x}\right)}{x^2 \cos\left(\frac{1}{x}\right)}, \quad q(x) = -\frac{1}{x^2 \cos\left(\frac{1}{x}\right)} \]

### Step 3: Find the Integrating Factor
The integrating factor \( \mu(x) \) is given by:

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

Calculating \( p(x) \):

\[ \int p(x) \, dx = \int -\frac{\sin\left(\frac{1}{x}\right)}{x^2 \cos\left(\frac{1}{x}\right)} \, dx \]

This integral can be simplified using the substitution \( u = \frac{1}{x} \), which gives \( dx = -\frac{1}{u^2} du \):

\[ \int -\frac{\sin(u)}{u^2 \cos(u)} \left(-\frac{1}{u^2}\right) du = \int \frac{\sin(u)}{u^2 \cos(u)} du \]

This leads to:

\[ \mu(x) = e^{\int -\tan\left(\frac{1}{x}\right) \, dx} = \sec\left(\frac{1}{x}\right) \]

### Step 4: Multiply the Equation by the Integrating Factor
Now, we multiply the entire differential equation by \( \sec\left(\frac{1}{x}\right) \):

\[ \sec\left(\frac{1}{x}\right) \frac{dy}{dx} - y \tan\left(\frac{1}{x}\right) = -\sec\left(\frac{1}{x}\right) \frac{1}{x^2} \]

### Step 5: Solve the Equation
This can be rewritten as:

\[ \frac{d}{dx}\left(y \sec\left(\frac{1}{x}\right)\right) = -\sec\left(\frac{1}{x}\right) \frac{1}{x^2} \]

Integrating both sides gives:

\[ y \sec\left(\frac{1}{x}\right) = \int -\sec\left(\frac{1}{x}\right) \frac{1}{x^2} \, dx + C \]

### Step 6: Evaluate the Integral
The integral can be evaluated using integration techniques, leading to:

\[ y \sec\left(\frac{1}{x}\right) = \tan\left(\frac{1}{x}\right) + C \]

### Step 7: Solve for \( y \)
Thus, we have:

\[ y = \tan\left(\frac{1}{x}\right) \cos\left(\frac{1}{x}\right) + C \cos\left(\frac{1}{x}\right) \]

### Step 8: Apply the Initial Condition
Given that \( y \to -1 \) as \( x \to \infty \), we can substitute \( x \to \infty \) (where \( \tan\left(\frac{1}{x}\right) \to 0 \) and \( \cos\left(\frac{1}{x}\right) \to 1 \)):

\[ -1 = 0 + C \cdot 1 \Rightarrow C = -1 \]

### Final Solution
Substituting \( C \) back into the equation gives:

\[ y = \tan\left(\frac{1}{x}\right) \cos\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) \]

This simplifies to:

\[ y = \frac{\sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)}{\cos\left(\frac{1}{x}\right)} \]

Thus, the final solution is:

\[ y = \frac{\sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)}{\cos\left(\frac{1}{x}\right)} \]