`x^2dy+(y-1/x)dx=0` at `y(1)=1` then find the value of `y(1/2)`

Answer

To solve the differential equation \( x^2 dy + \left( y - \frac{1}{x} \right) dx = 0 \) with the initial condition \( y(1) = 1 \), we can follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the given equation into a more standard form. We can write:

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

This simplifies to:

\[ \frac{dy}{dx} + \frac{y}{x^2} = \frac{1}{x^3} \]

### Step 2: Identifying \( p \) and \( q \)
From the rearranged equation, we can identify:

- \( p = \frac{1}{x^2} \)
- \( q = \frac{1}{x^3} \)

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

\[ \mu(x) = e^{\int p \, dx} = e^{\int \frac{1}{x^2} \, dx} \]

Calculating the integral:

\[ \int \frac{1}{x^2} \, dx = -\frac{1}{x} \]

Thus, the integrating factor is:

\[ \mu(x) = e^{-\frac{1}{x}} \]

### Step 4: Multiplying the Equation by the Integrating Factor
We multiply the entire differential equation by the integrating factor:

\[ e^{-\frac{1}{x}} \frac{dy}{dx} + e^{-\frac{1}{x}} \frac{y}{x^2} = e^{-\frac{1}{x}} \frac{1}{x^3} \]

### Step 5: Recognizing the Left Side as a Derivative
The left side can be recognized as the derivative of a product:

\[ \frac{d}{dx} \left( y e^{-\frac{1}{x}} \right) = e^{-\frac{1}{x}} \frac{1}{x^3} \]

### Step 6: Integrating Both Sides
Integrate both sides:

\[ \int \frac{d}{dx} \left( y e^{-\frac{1}{x}} \right) dx = \int e^{-\frac{1}{x}} \frac{1}{x^3} \, dx \]

The left side simplifies to:

\[ y e^{-\frac{1}{x}} = \int e^{-\frac{1}{x}} \frac{1}{x^3} \, dx + C \]

### Step 7: Solving the Integral on the Right Side
To solve the integral on the right side, we can use integration by parts or look it up. We find:

\[ \int e^{-\frac{1}{x}} \frac{1}{x^3} \, dx = -e^{-\frac{1}{x}} \left( \frac{1}{x^2} + C \right) \]

### Step 8: Applying the Initial Condition
Now we apply the initial condition \( y(1) = 1 \):

\[ 1 \cdot e^{-1} = -e^{-1} \left( \frac{1}{1^2} \right) + C \]

Solving for \( C \):

\[ e^{-1} = -e^{-1} + C \implies C = 2e^{-1} \]

### Step 9: Final Solution
Substituting back, we have:

\[ y e^{-\frac{1}{x}} = -e^{-\frac{1}{x}} \left( \frac{1}{x^2} \right) + 2e^{-1} \]

Multiplying through by \( e^{\frac{1}{x}} \):

\[ y = -\frac{1}{x^2} + 2e^{\frac{1}{x} - 1} \]

### Step 10: Finding \( y(1/2) \)
Now we substitute \( x = \frac{1}{2} \):

\[ y\left(\frac{1}{2}\right) = -\frac{1}{\left(\frac{1}{2}\right)^2} + 2e^{\frac{1}{\frac{1}{2}} - 1} = -4 + 2e^{2 - 1} = -4 + 2e \]

Thus, the final value is:

\[ y\left(\frac{1}{2}\right) = 2e - 4 \]