How many arbitrary constants are there in the particular solution of the differential equation `( dy )/( dx ) =- 4 xy^2 ,y(0) =1`

Answer

To determine how many arbitrary constants are present in the particular solution of the given differential equation, we can follow these steps:

### Step 1: Identify the Differential Equation
The given differential equation is:
\[ \frac{dy}{dx} = -4xy^2 \]

### Step 2: Recognize the Type of Differential Equation
This is a first-order ordinary differential equation (ODE).

### Step 3: Solve the Differential Equation
To solve the equation, we can separate the variables. Rearranging gives:
\[ \frac{dy}{y^2} = -4x \, dx \]

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

The left side integrates to:
\[ -\frac{1}{y} + C_1 \]
The right side integrates to:
\[ -2x^2 + C_2 \]

Thus, we have:
\[ -\frac{1}{y} = -2x^2 + C \]
where \(C = C_2 - C_1\) is a constant of integration.

### Step 5: Rearranging the Equation
Rearranging gives:
\[ \frac{1}{y} = 2x^2 + C \]

### Step 6: Apply the Initial Condition
We are given the initial condition \(y(0) = 1\). Substituting \(x = 0\) and \(y = 1\) into the equation:
\[ \frac{1}{1} = 2(0)^2 + C \implies 1 = C \]

### Step 7: Write the Particular Solution
Thus, the particular solution is:
\[ \frac{1}{y} = 2x^2 + 1 \]
or
\[ y = \frac{1}{2x^2 + 1} \]

### Step 8: Count the Arbitrary Constants
In the process of solving the differential equation, we introduced one arbitrary constant \(C\). However, since we applied the initial condition, we have determined the value of this constant.

### Conclusion
Therefore, the particular solution of the differential equation has **0 arbitrary constants** after applying the initial condition.

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