Step 1: Understanding the Concept:
This is a first-order, higher-degree differential equation. It's not in a standard form but appears to be solvable for \(x\). Another approach for multiple-choice questions is to verify the given options. We can take the proposed solution, find its corresponding differential equation, and check if it matches the given one.
Step 2: Key Formula or Approach:
1. Take the given solution from the options.
2. Differentiate it with respect to \(x\) to find an expression for \(p = \frac{dy}{dx}\).
3. Eliminate the constant \(c\) from the original solution and the differentiated equation.
4. The resulting equation in terms of \(x, y, p\) is the differential equation for the given family of curves. Compare it with the question's DE.
Step 3: Detailed Explanation:
Let's verify the correct option (A): \(y^2 = 2cx + c^3\).
This is the proposed family of solutions, where \(c\) is an arbitrary constant.
First, differentiate this equation with respect to \(x\):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(2cx + c^3) \]
\[ 2y \frac{dy}{dx} = 2c \]
Using the notation \(p = \frac{dy}{dx}\):
\[ 2yp = 2c \implies c = yp \]
Now we have an expression for the constant \(c\) in terms of \(y\) and \(p\). We can substitute this back into the original solution to eliminate \(c\).
Substitute \(c = yp\) into \(y^2 = 2cx + c^3\):
\[ y^2 = 2(yp)x + (yp)^3 \]
\[ y^2 = 2xyp + y^3p^3 \]
Since \(y\) is not always zero, we can divide the entire equation by \(y\):
\[ y = 2xp + y^2p^3 \]
Rearranging the terms to match the form in the question:
\[ y^2p^3 + 2xp - y = 0 \]
Wait, let me recheck the division. Division by y is valid if y is not identically zero.
Let's re-examine ‘y^2 = 2xyp + y^3p^3‘.
This rearranges to ‘y^3p^3 + 2xyp - y^2 = 0‘. Dividing by y: ‘y^2p^3 + 2xp - y = 0‘.
This is slightly different from the question ‘p^3y + 2px - y = 0‘.
Let's check my differentiation and substitution again.
Solution: \(y^2 = 2cx + c^3\).
Differentiate: \(2y \frac{dy}{dx} = 2c \implies yp=c\).
Substitute ‘c‘ into the solution:
\(y^2 = 2(yp)x + (yp)^3 = 2xyp + y^3p^3\).
So, \(y^2 - 2xyp - y^3p^3 = 0\).
Dividing by \(y\) (for \(y \neq 0\)) gives \(y - 2xp - y^2p^3 = 0\).
This is not matching the question ‘p^3y + 2px - y = 0‘.
Let's re-read the original equation: ‘p^3y + 2px - y = 0‘.
Let's check the signs. Maybe I made a mistake.
From the DE, let's solve for x: ‘2px = y - p^3y = y(1-p^3)‘. So ‘x = \frac{y(1-p^3)}{2p}‘. This is a form of Lagrange's equation, which is generally hard to solve.
Let's re-verify the substitution.
Option A: \(y^2 = 2cx + c^3\).
\(c = yp\).
Substitute into the DE: ‘p^3y + 2p(x) - y = 0‘. We need to eliminate ‘x‘ as well.
From the solution, \(2x = \frac{y^2-c^3}{c}\).
Substitute ‘x‘ and ‘c‘ into the DE:
‘p^3y + p(\frac{y^2-c^3}{c}) - y = 0‘.
Substitute ‘c = yp‘:
‘p^3y + p(\frac{y^2-(yp)^3}{yp}) - y = 0‘.
‘p^3y + \frac{y^2-y^3p^3}{y} - y = 0‘.
‘p^3y + (y - y^2p^3) - y = 0‘.
‘p^3y + y - y^2p^3 - y = 0‘.
‘p^3y - y^2p^3 = 0 \implies p^3y(1-y) = 0‘.
This does not work.
Let's try one more time. The process is to eliminate ‘c‘.
1) \(y^2 = 2cx + c^3\)
2) \(p = c/y\)
From (2), \(c = py\). Substitute this into (1):
\(y^2 = 2(py)x + (py)^3\)
\(y^2 = 2pxy + p^3y^3\)
Divide by y (assuming \(y \neq 0\)):
\(y = 2px + p^3y^2\)
This gives \(y^2p^3 + 2px - y = 0\).
Let's look at the question again: \(p^3y + 2px - y = 0\).
My derived equation is \(y^2p^3 + 2px - y = 0\).
The only difference is \(y^2p^3\) vs \(p^3y\).
This implies \(y^2p^3 = p^3y\), which means \(y^2=y\) or \(p=0\). This suggests that \(y=1\) or \(y=0\).
There seems to be a typo in the question or the solution. Let's assume the question was \(y^2p^3 + 2px - y = 0\). Then option A would be correct.
Given the context of such exams, typos are common. It is most likely that the question should have been \(y^2p^3 + 2px - y = 0\) or the solution should have led to \(p^3y + 2px - y = 0\).
Let's work backwards from the DE ‘p^3y + 2px - y = 0‘.
‘y(p^3 - 1) + 2px = 0‘.
This equation is a form of Clairaut's or Lagrange's equation. Let's rewrite it as ‘y = 2px/(1-p^3)‘.
This is a Lagrange's equation, solved by differentiating w.r.t x.
The provided solution ‘y^2=2cx+c^3‘ is the standard solution to a different DE. Let's assume the question is correct and there's a different way.
Given the MCQs, verification is the best bet.
‘y^2=2cx+c^3 \implies p=c/y‘.
The DE is ‘p^3y + 2px - y = 0‘.
Substitute ‘p=c/y‘:
‘(c/y)^3 y + 2(c/y)x - y = 0‘
‘c^3/y^2 + 2cx/y - y = 0‘
Multiply by ‘y^2‘:
‘c^3 + 2cxy - y^3 = 0‘.
We also have ‘y^2=2cx+c^3 \implies c^3 = y^2-2cx‘.
Substitute this into the derived relation:
‘(y^2 - 2cx) + 2cxy - y^3 = 0‘.
This must be an identity. It's not.
There is a clear mismatch. However, the form of the solution ‘y^2=2cx+c^3‘ is very specific. It's the general solution for ‘y = 2px + yp^3‘ (dividing by p... no this is not Clairaut's). It's the solution for ‘y = 2px + y^2p^3‘ (no, I derived this).
The problem seems flawed. However, if there's a typo and the DE was ‘-y p^3 + 2px + y = 0‘, then solving for ‘y‘ yields ‘y(1-p^3) = -2px‘, so ‘y = -2px/(1-p^3)‘.
Let's re-derive. Solution ‘y^2=2cx+c^3‘. Differentiate: ‘2y dy/dx = 2c‘, so ‘p=c/y‘.
Substitute ‘c=py‘ into solution: ‘y^2 = 2(py)x + (py)^3‘. ‘y = 2px + p^3y^2‘. This leads to ‘y - 2px - y^2 p^3=0‘.
Let's check the question again ‘p^3y + 2px - y = 0‘. This is ‘y = 2px + p^3y‘. This looks like my derived DE ‘y = 2px + p^3y^2‘ but with ‘y^2‘ replaced by ‘y‘. Let's assume the question is correct.
The structure of the question and option A suggests a Clairaut-type solution process where ‘c‘ is the parameter. Option A is famous as the solution to ‘y=2px+yp^2‘ not ‘p^3‘. It is possible that the powers are typos.
Given the situation, and that a specific answer is marked, there is likely a typo in the question and it should have been ‘y^2p^3 + 2px - y = 0‘. I will proceed assuming this.
Step 4: Final Answer:
Assuming the differential equation intended was \(y^2p^3 + 2px - y = 0\), the solution is indeed \(y^2 = 2cx + c^3\). The verification process shows that differentiating the solution and eliminating the constant \(c\) leads back to this corrected differential equation. Given the options, this is the most plausible interpretation.