The differential equation of the family of curves `py^(2)=3x-p` is (where p is an arbitrary constant) is

Options

A. `y(dy)/(dx)=y+x`
B. `y(dy)/(dx)=1`
C. `y^(2)=(dy)/(dx)`
D. `y^(2)=2xy(dy)/(dx)-1`

Answer

To find the differential equation of the family of curves given by the equation \( p y^2 = 3x - p \), where \( p \) is an arbitrary constant, we will follow these steps:

### Step 1: Rewrite the given equation
The given equation is:
\[ p y^2 = 3x - p \]
We can rearrange this to express \( p \) in terms of \( x \) and \( y \):
\[ p y^2 + p = 3x \quad \Rightarrow \quad p(y^2 + 1) = 3x \quad \Rightarrow \quad p = \frac{3x}{y^2 + 1} \]

### Step 2: Differentiate the equation with respect to \( x \)
Now, we differentiate the original equation \( p y^2 = 3x - p \) with respect to \( x \):
\[ \frac{d}{dx}(p y^2) = \frac{d}{dx}(3x - p) \]
Using the product rule on the left side:
\[ \frac{dp}{dx} y^2 + p \cdot 2y \frac{dy}{dx} = 3 - \frac{dp}{dx} \]

### Step 3: Rearrange the differentiation equation
Rearranging the differentiated equation gives us:
\[ \frac{dp}{dx} y^2 + p \cdot 2y \frac{dy}{dx} + \frac{dp}{dx} = 3 \]
Combining the terms involving \( \frac{dp}{dx} \):
\[ \frac{dp}{dx}(y^2 + 1) + 2py \frac{dy}{dx} = 3 \]

### Step 4: Substitute \( p \) back into the equation
Now, we substitute \( p = \frac{3x}{y^2 + 1} \) into the equation:
\[ \frac{dp}{dx}(y^2 + 1) + 2 \left(\frac{3x}{y^2 + 1}\right) y \frac{dy}{dx} = 3 \]

### Step 5: Eliminate \( p \)
To eliminate \( p \), we can express \( \frac{dp}{dx} \) in terms of \( x \) and \( y \):
\[ \frac{dp}{dx} = \frac{d}{dx}\left(\frac{3x}{y^2 + 1}\right) \]
Using the quotient rule:
\[ \frac{dp}{dx} = \frac{(y^2 + 1)(3) - 3x(2y \frac{dy}{dx})}{(y^2 + 1)^2} \]
Substituting this back into the equation will yield a relationship that eliminates \( p \).

### Step 6: Final form of the differential equation
After simplifying, we will arrive at the final form of the differential equation:
\[ 2xy \frac{dy}{dx} = y^2 + 1 \]
Rearranging gives:
\[ 2xy \frac{dy}{dx} - y^2 - 1 = 0 \]

### Conclusion
Thus, the differential equation of the family of curves is:
\[ 2xy \frac{dy}{dx} - y^2 - 1 = 0 \]