The differential equation of the family of curves `cy ^(2) =2x +c` (where c is an arbitrary constant.) is:

Options

A. `(xdy)/(dx)=1`
B. `((dy)/(dx)) =(2xly)/(dx )+1`
C. `y ^(2)=2xy(dy)/(dx)+1`
D. `y ^(2) =(2ydy )/(dx)+1`

Answer

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

### Step 1: Differentiate the given equation with respect to \( x \)

The given equation is:
\[ c y^2 = 2x + c \]

Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(c y^2) = \frac{d}{dx}(2x + c) \]

Using the product rule on the left side:
\[ c \cdot 2y \frac{dy}{dx} = 2 + 0 \]
This simplifies to:
\[ 2c y \frac{dy}{dx} = 2 \]

### Step 2: Solve for \( c \)

From the equation \( 2c y \frac{dy}{dx} = 2 \), we can isolate \( c \):
\[ c = \frac{1}{y \frac{dy}{dx}} \]

### Step 3: Substitute \( c \) back into the original equation

Now, substitute the expression for \( c \) back into the original equation:
\[ \frac{1}{y \frac{dy}{dx}} y^2 = 2x + \frac{1}{y \frac{dy}{dx}} \]

Multiplying through by \( y \frac{dy}{dx} \) to eliminate the fraction:
\[ y^2 = 2xy \frac{dy}{dx} + 1 \]

### Step 4: Rearranging the equation

This gives us the differential equation:
\[ y^2 - 1 = 2xy \frac{dy}{dx} \]

### Final Result

Thus, the differential equation of the family of curves \( c y^2 = 2x + c \) is:
\[ y^2 = 2xy \frac{dy}{dx} + 1 \]

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