The differential equation of the family of curves, `x^(2)=4b(y+b),b in R,` is :

Options

A. `x(y')^(2)=x+2yy'`
B. `x(y')^(2)=2y y'-x`
C. `x(y')^(2)=x-2 y y'`
D. `xy''=y'

Answer

To find the differential equation of the family of curves given by the equation \( x^2 = 4b(y + b) \) where \( b \in \mathbb{R} \), we can follow these steps:

### Step 1: Rewrite the given equation
Start with the equation:
\[ x^2 = 4b(y + b) \]

### Step 2: Expand the equation
Expand the right-hand side:
\[ x^2 = 4by + 4b^2 \]

### Step 3: Differentiate with respect to \( x \)
Differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(x^2) = \frac{d}{dx}(4by + 4b^2) \]
This gives:
\[ 2x = 4b \frac{dy}{dx} + 0 \]
(Here, the derivative of \( 4b^2 \) with respect to \( x \) is zero since \( b \) is treated as a constant.)

### Step 4: Solve for \( b \)
From the differentiated equation, we can isolate \( b \):
\[ 2x = 4b \frac{dy}{dx} \implies b = \frac{2x}{4 \frac{dy}{dx}} = \frac{x}{2 \frac{dy}{dx}} \]

### Step 5: Substitute \( b \) back into the original equation
Substituting \( b \) back into the original equation:
\[ x^2 = 4\left(\frac{x}{2 \frac{dy}{dx}}\right)(y + \frac{x}{2 \frac{dy}{dx}}) \]
This simplifies to:
\[ x^2 = 2x(y + \frac{x}{2 \frac{dy}{dx}}) \]

### Step 6: Simplify the equation
Distributing on the right side:
\[ x^2 = 2xy + \frac{x^2}{\frac{dy}{dx}} \]
Now, multiply both sides by \( \frac{dy}{dx} \) to eliminate the fraction:
\[ x^2 \frac{dy}{dx} = 2xy \frac{dy}{dx} + x^2 \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ x^2 \frac{dy}{dx} - 2xy \frac{dy}{dx} = x^2 \]
Factoring out \( \frac{dy}{dx} \):
\[ \frac{dy}{dx}(x^2 - 2xy) = x^2 \]

### Step 8: Final form of the differential equation
Thus, we can express the differential equation as:
\[ x \frac{dy}{dx} = 2y + x \]
or equivalently:
\[ x \frac{dy}{dx} = x + 2y \]

### Conclusion
The differential equation of the family of curves is:
\[ x \frac{dy}{dx} = x + 2y \]