The differential equation of the family of curves represented by `y^3= cx+ c^3+ c^2-1`,where c is an arbitrary constant is of :

Options

A. Order 1, degree 1
B. order 2, degree 1
C. order 1, degree 3
D. order 2, degree2

Answer

To find the differential equation of the family of curves represented by the equation \( y^3 = cx + c^3 + c^2 - 1 \), where \( c \) is an arbitrary constant, we will follow these steps:

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

We start with the equation:
\[ y^3 = cx + c^3 + c^2 - 1 \]

Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(y^3) = \frac{d}{dx}(cx + c^3 + c^2 - 1) \]

Using the chain rule on the left side:
\[ 3y^2 \frac{dy}{dx} = c \]

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

From the equation \( 3y^2 \frac{dy}{dx} = c \), we can express \( c \) in terms of \( y \) and \( \frac{dy}{dx} \):
\[ c = 3y^2 \frac{dy}{dx} \]

### Step 3: Differentiate again to eliminate \( c \)

Now, we will differentiate the expression \( c = 3y^2 \frac{dy}{dx} \) with respect to \( x \):
\[ \frac{d}{dx}(c) = \frac{d}{dx}(3y^2 \frac{dy}{dx}) \]

Since \( c \) is a constant, its derivative is 0:
\[ 0 = \frac{d}{dx}(3y^2 \frac{dy}{dx}) \]

Using the product rule on the right side:
\[ 0 = 3 \left(2y \frac{dy}{dx} \frac{dy}{dx} + y^2 \frac{d^2y}{dx^2}\right) \]

This simplifies to:
\[ 0 = 6y \left(\frac{dy}{dx}\right)^2 + 3y^2 \frac{d^2y}{dx^2} \]

### Step 4: Factor out common terms

We can factor out \( 3y \):
\[ 0 = 3y \left(2 \left(\frac{dy}{dx}\right)^2 + y \frac{d^2y}{dx^2}\right) \]

### Step 5: Set the equation to find the order and degree

For the differential equation, we have:
\[ 2 \left(\frac{dy}{dx}\right)^2 + y \frac{d^2y}{dx^2} = 0 \]

The highest order derivative is \( \frac{d^2y}{dx^2} \), which indicates that the order of the differential equation is 2. The degree of the equation is 1, as it is a polynomial in \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \).

### Final Conclusion

Thus, the differential equation of the family of curves is of order 2 and degree 1.