Solution of the diff. eqn. `(dy)/(dx) +(1)/(x) =(e^(y))/(x^(2))` is ………..

Answer

To solve the differential equation

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

we will follow these steps:

### Step 1: Rearranging the Equation
First, we rearrange the equation to isolate the exponential term on one side:

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

### Step 2: Substitution
Next, we make the substitution \( v = e^{-y} \). Then, we differentiate \( v \):

\[ \frac{dv}{dx} = -e^{-y} \frac{dy}{dx} = -v \frac{dy}{dx}. \]

Substituting \( e^{-y} = v \) into the equation gives:

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

Rearranging, we have:

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

### Step 3: Identifying the Linear Differential Equation
This is a linear first-order differential equation in the form:

\[ \frac{dv}{dx} + P(x)v = Q(x), \]

where \( P(x) = \frac{1}{x} \) and \( Q(x) = -\frac{1}{x^2} \).

### Step 4: Finding the Integrating Factor
The integrating factor \( \mu(x) \) is given by:

\[ \mu(x) = e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\ln |x|} = |x|. \]

### Step 5: Multiplying through by the Integrating Factor
We multiply the entire equation by the integrating factor:

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

This simplifies to:

\[ |x| \frac{dv}{dx} + v = -\frac{1}{x}. \]

### Step 6: Integrating
Now we integrate both sides:

\[ \int \left( |x| \frac{dv}{dx} + v \right) dx = \int -\frac{1}{x} dx. \]

The left side becomes:

\[ v|x| = -\ln |x| + C, \]

where \( C \) is the constant of integration.

### Step 7: Substituting Back
Now we substitute back \( v = e^{-y} \):

\[ e^{-y}|x| = -\ln |x| + C. \]

### Step 8: Solving for \( y \)
Rearranging gives:

\[ e^{-y} = \frac{-\ln |x| + C}{|x|}. \]

Taking the natural logarithm on both sides:

\[ -y = \ln\left(\frac{-\ln |x| + C}{|x|}\right), \]

which leads to:

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

### Final Step: Simplifying the Expression
Thus, we can express the solution as:

\[ y = \ln |x| + \ln(-\ln |x| + C). \]

### Final Answer
The solution of the differential equation is:

\[ y = \ln |x| + \ln(-\ln |x| + C). \]

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