Solve the following differential equation: `(dy)/(dx)=(1+y^2)/(y^3)`

Explanation

To solve the differential equation \(\frac{dy}{dx} = \frac{1 + y^2}{y^3}\), we will follow these steps:

### Step 1: Rewrite the equation
The given differential equation can be rewritten as:
\[ \frac{dy}{dx} = \frac{1 + y^2}{y^3} \]
This indicates that the right-hand side is a function of \(y\) only.

### Step 2: Separate variables
We can separate the variables by taking the reciprocal of both sides:
\[ \frac{dx}{dy} = \frac{y^3}{1 + y^2} \]

### Step 3: Integrate both sides
Now, we will integrate both sides:
\[ \int dx = \int \frac{y^3}{1 + y^2} dy \]
The left side integrates to:
\[ x = \int \frac{y^3}{1 + y^2} dy + C \]
where \(C\) is the constant of integration.

### Step 4: Simplify the integral on the right side
To integrate \(\frac{y^3}{1 + y^2}\), we can break it down:
\[ \frac{y^3}{1 + y^2} = \frac{y^3}{1 + y^2} = y^3 \cdot \frac{1}{1 + y^2} \]
We can rewrite \(y^3\) as \(y^2 \cdot y\):
\[ \int \frac{y^3}{1 + y^2} dy = \int \frac{y^2 \cdot y}{1 + y^2} dy \]

### Step 5: Use substitution
Let \(t = 1 + y^2\), then \(dt = 2y dy\) or \(y dy = \frac{1}{2} dt\). Thus, we can express \(y^2\) in terms of \(t\):
\[ y^2 = t - 1 \]
Now substituting into the integral:
\[ \int \frac{y^2 \cdot y}{1 + y^2} dy = \int \frac{(t - 1)}{t} \cdot \frac{1}{2} dt \]
This simplifies to:
\[ \frac{1}{2} \int \left(1 - \frac{1}{t}\right) dt = \frac{1}{2} \left(t - \log |t|\right) + C \]

### Step 6: Substitute back
Now substituting back \(t = 1 + y^2\):
\[ x = \frac{1}{2} \left((1 + y^2) - \log |1 + y^2|\right) + C \]

### Final Step: Rearranging the equation
To express the solution clearly:
\[ x = \frac{1 + y^2}{2} - \frac{1}{2} \log |1 + y^2| + C \]

### Summary of the solution
The solution to the differential equation \(\frac{dy}{dx} = \frac{1 + y^2}{y^3}\) is:
\[ x = \frac{1 + y^2}{2} - \frac{1}{2} \log |1 + y^2| + C \]