The equation of the curve satisfying the differential equation `(dy)/(dx)+2(y)/(x^(2))=(2)/(x^(2))` and passing through `((1)/(2),e^(4)+1)` is

Options

A. `y=e^(2x)+1`
B. `y=e^((2)/(x))-1`
C. `y=1+e^((2)/(x))`
D. `y=1+e^(-x)`

Answer

To solve the differential equation

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

with the initial condition that the curve passes through the point \(\left(\frac{1}{2}, e^4 + 1\right)\), we will follow these steps:

### Step 1: Rewrite the Differential Equation
Multiply the entire equation by \(x^2\) to eliminate the fractions:

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

### Step 2: Rearrange the Equation
Rearranging gives us:

\[ x^2 \frac{dy}{dx} = 2 - 2y \]

### Step 3: Separate Variables
We can separate the variables by moving all terms involving \(y\) to one side and \(x\) to the other:

\[ \frac{dy}{2 - 2y} = \frac{dx}{x^2} \]

### Step 4: Integrate Both Sides
Now, we integrate both sides:

\[ \int \frac{dy}{2 - 2y} = \int \frac{dx}{x^2} \]

The left side integrates to:

\[ -\ln|2 - 2y| = -\frac{1}{x} + C \]

### Step 5: Solve for \(y\)
Rearranging gives us:

\[ \ln|2 - 2y| = \frac{1}{x} - C \]

Exponentiating both sides:

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

Let \(k = e^{-C}\), then:

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

Solving for \(y\):

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

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

### Step 6: Apply the Initial Condition
Now we apply the initial condition \(y\left(\frac{1}{2}\right) = e^4 + 1\):

\[ e^4 + 1 = 1 - \frac{k}{2e^{2}} \]

This simplifies to:

\[ e^4 = -\frac{k}{2e^{2}} \]

Thus:

\[ k = -2e^{6} \]

### Step 7: Substitute \(k\) Back
Substituting \(k\) back into our equation for \(y\):

\[ y = 1 + \frac{e^{6}}{e^{\frac{1}{x}}} \]

This simplifies to:

\[ y = 1 + e^{6 - \frac{1}{x}} \]

### Final Step: Write the Final Equation
Thus, the equation of the curve is:

\[ y = 1 + e^{6 - \frac{1}{x}} \]