Solve each of the following differential equations
`(xy^2+x)dx+(yx^2+y)dy=0 , y(0)=1`.

Answer

To solve the differential equation \( (xy^2 + x)dx + (yx^2 + y)dy = 0 \) with the initial condition \( y(0) = 1 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ (xy^2 + x)dx + (yx^2 + y)dy = 0 \]
We can rearrange this as:
\[ yx^2 + y \, dy = - (xy^2 + x) \, dx \]

### Step 2: Factor out common terms
Factoring out \(y\) from the left side and \(x\) from the right side gives us:
\[ y(x^2 + 1) \, dy = -x(y^2 + 1) \, dx \]

### Step 3: Separate variables
Now we separate the variables \(y\) and \(x\):
\[ \frac{y \, dy}{y^2 + 1} = -\frac{x \, dx}{x^2 + 1} \]

### Step 4: Integrate both sides
Next, we integrate both sides:
\[ \int \frac{y \, dy}{y^2 + 1} = -\int \frac{x \, dx}{x^2 + 1} \]

Using the substitution \( t = y^2 + 1 \) for the left side and \( m = x^2 + 1 \) for the right side, we have:
- For the left side:
\[ dt = 2y \, dy \implies y \, dy = \frac{dt}{2} \]
Thus,
\[ \int \frac{y \, dy}{y^2 + 1} = \frac{1}{2} \int \frac{dt}{t} = \frac{1}{2} \ln |t| + C_1 = \frac{1}{2} \ln(y^2 + 1) + C_1 \]

- For the right side:
\[ dm = 2x \, dx \implies x \, dx = \frac{dm}{2} \]
Thus,
\[ -\int \frac{x \, dx}{x^2 + 1} = -\frac{1}{2} \int \frac{dm}{m} = -\frac{1}{2} \ln |m| + C_2 = -\frac{1}{2} \ln(x^2 + 1) + C_2 \]

### Step 5: Combine the results
Combining both integrals gives us:
\[ \frac{1}{2} \ln(y^2 + 1) + \frac{1}{2} \ln(x^2 + 1) = C \]
We can express this as:
\[ \ln(y^2 + 1) + \ln(x^2 + 1) = 2C \]
Using the property of logarithms:
\[ \ln((y^2 + 1)(x^2 + 1)) = 2C \]
Let \( K = e^{2C} \), we have:
\[ (y^2 + 1)(x^2 + 1) = K \]

### Step 6: Apply the initial condition
Using the initial condition \( y(0) = 1 \):
\[ (1^2 + 1)(0^2 + 1) = K \implies 2 \cdot 1 = K \implies K = 2 \]
Thus, our equation becomes:
\[ (y^2 + 1)(x^2 + 1) = 2 \]

### Step 7: Final form
We can rewrite this as:
\[ y^2 + 1 = \frac{2}{x^2 + 1} \]
This is the solution to the differential equation.

### Summary of the solution:
The solution to the differential equation \( (xy^2 + x)dx + (yx^2 + y)dy = 0 \) with the initial condition \( y(0) = 1 \) is:
\[ (y^2 + 1)(x^2 + 1) = 2 \]