The differential equation representing the family of curves \(\mathrm{y}^{2}=2 \mathrm{c}(\mathrm{x}+\mathrm{v} \mathrm{c})\), where \(\mathrm{c}\) is a positive parameter, is of order and degree as follows. (A) order 2, degree 1 (B) order 1 , degree 2 (C) order 2 , degree 2 (D) order 1 , degree 3

The given equation is \(y^2=2c(x+vc)\). Here \(v\) and \(c\) are arbitrary constants. We need to eliminate these constants from the equation to express it as a differential equation. First replace \(c\) using \(y^2/2(x+vc)\) and differentiate the resulting equation with respect to \(x\) to eliminate \(v\).

Start with the equation \(c = \frac{y^2}{2(x+vc)}\). Differentiate this equation with respect to \(x\) using chain rule: \(\frac{d}{dx}c = \frac{d}{dx}[y^2/2(x+vc)]\) Since \(c\) is a constant, it's differentiation will be zero and on the other side of the equation using quotient and chain rule: \(0=\frac{[(2yy')(2(x+vc)) - y^2(2+2vc')]}{4(x+vc)^2}\) Considering that the derivative of \(c\) with respect to \(x\) is zero (\(c' = 0\)), the equation simplifies to: \(0=\frac{(2yy')(2x+2vc) - 2y^2}{4(x+vc)^2}\) Then after simplifying it we get: \(0=(yy'[x+vc] - y^2)/(x+vc)^2\)

The revised equation is a differential equation. We can see the order and degree of this differential equation. Order of a differential equation is basically the order of the highest order derivative present in the equation. Degree is the power of the highest order derivative in the equation only if the differential coefficients in the equation are free from radicals and fractions. Here in \(0=(yy'[x+vc] - y^2)/(x+vc)^2\), the highest order of derivative is 1 (which is \(y'\)). The power of this highest order derivative is also 1 since it is not under any radical or fractional power. Hence, the order of the differential equation is 1 and the degree is 1. So, none of the given options A, B, C, D are correct.