Solve the following differential equation: `(dy)/(dx)+(1+y^2)/y=0`

Explanation

To solve the differential equation

\[ \frac{dy}{dx} + \frac{1+y^2}{y} = 0, \]

we can follow these steps:

### Step 1: Rearranging the Equation
First, we can rearrange the equation to isolate \(\frac{dy}{dx}\):

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

### Step 2: Separating Variables
Next, we can separate the variables by rewriting the equation in terms of \(dx\) and \(dy\):

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

### Step 3: Integrating Both Sides
Now, we integrate both sides. The left side integrates to \(x\), and for the right side, we need to integrate \(-\frac{y}{1+y^2}\):

\[ \int dx = -\int \frac{y}{1+y^2} dy. \]

The integral on the right can be simplified by using a substitution. Let \(t = 1 + y^2\), then \(dt = 2y \, dy\) or \(\frac{dt}{2} = y \, dy\). Thus, we have:

\[ -\int \frac{y}{1+y^2} dy = -\frac{1}{2} \int \frac{1}{t} dt = -\frac{1}{2} \ln |t| + C, \]

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

### Step 4: Back Substituting
Now we substitute back \(t = 1 + y^2\):

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

### Step 5: Final Form
To express the solution more clearly, we can rearrange it:

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

Exponentiating both sides gives:

\[ |1+y^2| = e^{2(C-x)}. \]

Thus, the final solution can be expressed as:

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

where \(K = e^{2C}\) is a constant.

### Summary of the Solution
The solution to the differential equation is:

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

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