The differential equation `y=px+f(p)`, …………..(i) where `p=(dy)/(dx)`,is known as Clairout's equation. To solve equation i) differentiate it with respect to x, which gives either `(dp)/(dx)=0 rArr p =c`………….(ii) or `x+f^(i)(p)=0`…………(iii) Which of the following is true about solutions of differential equation `y=xy^(')+sqrt(1+y^('2))`?

Options

A. the general solution of equation is family of parabolas
B. the general solution of equation is family of circles
C. the singular solution of equation is circle
D. the singular solution of equation is ellipse

Explanation

To solve the differential equation given by \( y = px + f(p) \) where \( p = \frac{dy}{dx} \), we will follow the steps outlined in the video transcript.

### Step 1: Differentiate the equation with respect to \( x \)
Starting with the equation:
\[ y = px + f(p) \]
we differentiate both sides with respect to \( x \):
\[ \frac{dy}{dx} = p + x \frac{dp}{dx} + \frac{df}{dp} \cdot \frac{dp}{dx} \]
This simplifies to:
\[ p = p + x \frac{dp}{dx} + f'(p) \frac{dp}{dx} \]

### Step 2: Rearranging the equation
We can rearrange the equation to isolate \( \frac{dp}{dx} \):
\[ 0 = x \frac{dp}{dx} + f'(p) \frac{dp}{dx} \]
Factoring out \( \frac{dp}{dx} \):
\[ 0 = \frac{dp}{dx} (x + f'(p)) \]

### Step 3: Setting the factors to zero
From the equation above, we have two cases:
1. \( \frac{dp}{dx} = 0 \) which implies \( p = c \) (a constant).
2. \( x + f'(p) = 0 \).

### Step 4: Analyzing the second case
From the second case, we can express \( f'(p) \) as:
\[ f'(p) = -x \]

### Step 5: Finding the general solution
Now, we need to analyze the given differential equation:
\[ y = xy' + \sqrt{1 + (y')^2} \]
Let \( p = y' \). We can rewrite the equation as:
\[ y = xp + \sqrt{1 + p^2} \]

### Step 6: Differentiating with respect to \( x \)
Differentiating both sides with respect to \( x \):
\[ \frac{dy}{dx} = p + x \frac{dp}{dx} + \frac{1}{2\sqrt{1 + p^2}} \cdot 2p \frac{dp}{dx} \]
This simplifies to:
\[ p = p + x \frac{dp}{dx} + \frac{p}{\sqrt{1 + p^2}} \frac{dp}{dx} \]

### Step 7: Rearranging again
Rearranging gives:
\[ 0 = x \frac{dp}{dx} + \frac{p}{\sqrt{1 + p^2}} \frac{dp}{dx} \]
Factoring out \( \frac{dp}{dx} \):
\[ 0 = \frac{dp}{dx} \left( x + \frac{p}{\sqrt{1 + p^2}} \right) \]

### Step 8: Setting the factors to zero
This leads us to the same two cases:
1. \( \frac{dp}{dx} = 0 \) which gives \( p = c \).
2. \( x + \frac{p}{\sqrt{1 + p^2}} = 0 \).

### Step 9: Finding the singular solution
From the second case, we can solve for \( p \):
\[ \frac{p}{\sqrt{1 + p^2}} = -x \]
Squaring both sides gives:
\[ p^2 = x^2(1 + p^2) \]
This leads to:
\[ p^2(1 - x^2) = x^2 \]
Thus, we can express \( p \) as:
\[ p = \frac{x}{\sqrt{1 - x^2}} \]

### Step 10: General solution
Substituting back to find \( y \):
\[ y = cx + \sqrt{1 + c^2} \]
This leads us to the general solution:
\[ x^2 + y^2 = 1 \]
This represents a circle.

### Conclusion
The correct answer regarding the solutions of the differential equation \( y = xy' + \sqrt{1 + (y')^2} \) is that it represents a singular solution of the equation, which is a circle.

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