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The differential equation by eliminating arbitrary constant \(a\) from :

\[ y = a(x - a)^{2} \]

is :

  • A.\(\left(\dfrac{dy}{dx}\right)^{3} - 4xy\dfrac{dy}{dx} + 8y = 0\)
  • B.\(\left(\dfrac{dy}{dx}\right)^{3} - 4xy\dfrac{dy}{dx} + 8y^{2} = 0\)
  • C.\(\left(\dfrac{dy}{dx}\right)^{2} - 4xy\dfrac{dy}{dx} + 8y^{2} = 0\)
  • D.\(\left(\dfrac{dy}{dx}\right)^{2} - 4xy\dfrac{dy}{dx} + 8y = 0\)

Answer: B

The equation is \((y')^{3} - 4xyy' + 8y^{2} = 0\).

Write \(p = dy/dx\). From \(y = a(x - a)^{2}\), \(p = 2a(x - a)\). Squaring, \(p^{2} = 4a^{2}(x - a)^{2} = 4ay\), so \(a = p^{2}/(4y)\). Also \(x - a = p/(2a) = 2y/p\), so \(x = p^{2}/(4y) + 2y/p\). Multiply by \(4yp\):

\[ \begin{aligned} 4xyp &= p^{3} + 8y^{2} \\ p^{3} - 4xyp + 8y^{2} &= 0 \end{aligned} \]

The equation is of degree 3 in \(p\), which rules out the two squared options, and the constant term is \(8y^{2}\), not \(8y\).

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