The differential equation which represents the family of curves \(\mathrm{y}=\mathrm{c}_{1} \mathrm{e}^{(\mathrm{c}) 2 \mathrm{x}}\), where \(\mathrm{c}_{1}\) and \(\mathrm{c}_{2}\) are arbitarary constants, is: (A) \(y^{\prime}=y^{2}\) (B) \(\mathrm{y}^{\prime \prime}=\mathrm{y}^{\prime} \mathrm{y}\) (C) \(\mathrm{yy}^{\prime \prime}=\left(\mathrm{y}^{1}\right)^{2}\) (D) \(\mathrm{yy}^{\prime \prime}=\mathrm{y}^{\prime}\)

Differentiate the given equation y = c_1 e^{c_2 2x} with respect to x to find the first derivative: \(y' = \frac{dy}{dx} = c_1 (2c_2) e^{c_2 2x}\)

Differentiate \(y'\) with respect to x to find the second derivative: \(y'' = \frac{d^2y}{dx^2} = c_1 (2c_2)^2 e^{c_2 2x}\)

We have: \(y = c_1 e^{c_2 2x}\) \(y' = c_1 (2c_2) e^{c_2 2x}\) \(y'' = c_1 (2c_2)^2 e^{c_2 2x}\) To eliminate c_1 and c_2, we can use the second equation and divide it by the first equation and then divide the third equation by the second equation. \(\frac{y'}{y} = 2c_2\) \(\frac{y''}{y'} = 2c_2\) Since \(\frac{y'}{y} = \frac{y''}{y'}\), we can rewrite the equation as: \(y' y = y''y\) Now, we check our options: (A) \(y' = y^2\) This is not a match because there is no \(y^2\) term in our transformed equation. (B) \(y''= y' y\) This is not a match because there is a product term missing from our transformed equation. (C) \(y y''= (y^1)^2\) This is a match with our transformed equation \(y' y = y''y\)! (D) \(yy''= y'\) This is not a match because our transformed equation has equal signs on both sides.

Based on our transformed differential equation and the given options, we can conclude that the differential equation which represents the family of curves \(y = c_1 e^{c_2 2x}\) is: (C) \(y y''= (y^1)^2\)