The solution of the differential equation `x^2(dy)/(dx)cos(1/x)-ysin(1/x)=-1,` where `y--1` as`x -oo` is

Options

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

Explanation

To solve the 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 will follow these steps:

### Step 1: Rewrite the Differential Equation
We start by rewriting the equation in a more standard form. We can divide through 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 standard form of a linear differential equation \( \frac{dy}{dx} + p(x)y = q(x) \), we identify:

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

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

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

Calculating the integral:

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

Let \( t = \frac{1}{x} \), then \( dx = -\frac{1}{t^2} dt \) and \( x^2 = \frac{1}{t^2} \):

\[ \int -\frac{\sin(t)}{\frac{1}{t^2} \cos(t)} \left(-\frac{1}{t^2}\right) dt = \int \frac{\sin(t)}{\cos(t)} dt = \int \tan(t) dt = -\log|\cos(t)| + C. \]

Thus, the integrating factor is:

\[ \mu(x) = e^{-\log|\cos\left(\frac{1}{x}\right)|} = \sec\left(\frac{1}{x}\right). \]

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

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

### Step 5: Solve the Equation
The left-hand side can be expressed as the derivative of a product:

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

Integrating both sides gives:

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

### Step 6: Evaluate the Integral
Using the substitution \( t = \frac{1}{x} \):

\[ \int -\sec(t) dt = -\ln|\sec(t) + \tan(t)| + C. \]

Thus, we have:

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

### Step 7: Solve for \( y \)
Now, we can isolate \( y \):

\[ y = -\ln\left|\sec\left(\frac{1}{x}\right) + \tan\left(\frac{1}{x}\right)\right| \cos\left(\frac{1}{x}\right) + C \cos\left(\frac{1}{x}\right). \]

### Step 8: Apply the Initial Condition
As \( x \to \infty \), \( \sec\left(\frac{1}{x}\right) \to 1 \) and \( \tan\left(\frac{1}{x}\right) \to 0 \), thus:

\[ y \to -\ln(1) + C \cdot 1 = C. \]

Given \( y \to -1 \) as \( x \to \infty \), we find \( C = -1 \).

### Final Solution
Substituting \( C \) back, we have:

\[ y = -\ln\left|\sec\left(\frac{1}{x}\right) + \tan\left(\frac{1}{x}\right)\right| \cos\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right). \]