Eliminate the arbitrary constant in the equation y? = 4ax. а. х?dу + 2хуdx —D 0 b. x?dy – 2xydx = 0 c. y?dx + 2xydy = 0 d. y?dx – 2xydy = 0

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Polynomials

Try substituting 1. You will get 6 and not 0. Try other real values also. What About -2? Let us check.