Solve the equation `(dy)/(dx)=1/x=(e^y)/(x^2)`

Explanation

To solve the differential equation

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

we can start by rewriting the equation in a more manageable form.

### Step 1: Rewrite the equation
We can express the equation as:

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

### Step 2: Multiply through by \( e^{-y} \)
To simplify the equation, we multiply through by \( e^{-y} \):

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

### Step 3: Let \( v = e^{-y} \)
Now, let \( v = e^{-y} \). Then, differentiating both sides gives us:

\[ \frac{dv}{dx} = -e^{-y} \frac{dy}{dx} \quad \Rightarrow \quad e^{-y} \frac{dy}{dx} = -\frac{dv}{dx} \]

### Step 4: Substitute into the equation
Substituting this back into our equation gives:

\[ -\frac{dv}{dx} + \frac{v}{x} = \frac{1}{x^2} \]

Rearranging this, we have:

\[ \frac{dv}{dx} - \frac{v}{x} = -\frac{1}{x^2} \]

### Step 5: Identify the integrating factor (IF)
To solve this linear differential equation, we need to find the integrating factor (IF). The IF is given by:

\[ IF = e^{\int -\frac{1}{x} \, dx} = e^{-\ln x} = \frac{1}{x} \]

### Step 6: Multiply through by the integrating factor
Now, we multiply the entire equation by the integrating factor:

\[ \frac{1}{x} \frac{dv}{dx} - \frac{v}{x^2} = -\frac{1}{x^3} \]

This can be rewritten as:

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

### Step 7: Integrate both sides
Now, we integrate both sides:

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

The left side simplifies to:

\[ \frac{v}{x} = \frac{1}{2x^2} + C \]

### Step 8: Solve for \( v \)
Multiplying through by \( x \):

\[ v = \frac{1}{2x} + Cx \]

### Step 9: Substitute back for \( y \)
Recall that \( v = e^{-y} \), so we have:

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

Taking the natural logarithm of both sides gives:

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

Thus,

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

### Final Solution
The final solution to the differential equation is:

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