Solve the following initial value problems and find the corresponding solution curves :
`y'+2y^(2)=0, y(1)=1`

Answer

To solve the initial value problem \( y' + 2y^2 = 0 \) with the initial condition \( y(1) = 1 \), we follow these steps:

### Step 1: Rewrite the differential equation
The given equation can be rewritten as:
\[ y' = -2y^2 \]

### Step 2: Separate the variables
We can separate the variables by rearranging the equation:
\[ \frac{dy}{2y^2} = -dx \]

### Step 3: Integrate both sides
Now, we integrate both sides:
\[ \int \frac{dy}{2y^2} = \int -dx \]
The left side can be simplified:
\[ \frac{1}{2} \int y^{-2} dy = -\int dx \]
Integrating gives:
\[ \frac{1}{2} \left( -\frac{1}{y} \right) = -x + C \]
This simplifies to:
\[ -\frac{1}{2y} = -x + C \]

### Step 4: Solve for \( y \)
Multiplying through by -2 gives:
\[ \frac{1}{y} = 2x - 2C \]
Let \( c = -2C \), then:
\[ \frac{1}{y} = 2x + c \]
Thus, we have:
\[ y = \frac{1}{2x + c} \]

### Step 5: Use the initial condition to find \( c \)
Now we apply the initial condition \( y(1) = 1 \):
\[ 1 = \frac{1}{2(1) + c} \]
This simplifies to:
\[ 1 = \frac{1}{2 + c} \]
Cross-multiplying gives:
\[ 2 + c = 1 \implies c = -1 \]

### Step 6: Write the final solution
Substituting \( c \) back into the equation for \( y \):
\[ y = \frac{1}{2x - 1} \]

### Final Result
The solution curve for the initial value problem is:
\[ y = \frac{1}{2x - 1}, \quad \text{where } x \neq \frac{1}{2} \]