Formation Of Differential Equations

Question

Given the family of curves y = A\cos x + B\sin x with two arbitrary constants, form the differential equation by eliminating both parameters completely.

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Forming a differential equation from a family of curves means eliminating the arbitrary constants by differentiating as many times as there are constants. Since y = A\cos x + B\sin x contains two independent constants A and B, we must differentiate twice to remove both. First, \frac{dy}{dx} = -A\sin x + B\cos x, which still carries the constants. Differentiating again gives \frac{d^2y}{dx^2} = -A\cos x - B\sin x. The crucial observation is that the right side is exactly the negative of the original y, since -A\cos x - B\sin x = -(A\cos x + B\sin x) = -y. Therefore \frac{d^2y}{dx^2} = -y, which rearranges to \frac{d^2y}{dx^2} + y = 0. Option \frac{d^2y}{dx^2} - y = 0 corresponds to the family A e^x + B e^{-x}, not trigonometric. Option \frac{dy}{dx} + y = 0 eliminates only one constant and is first order. Option \frac{d^2y}{dx^2} + \frac{dy}{dx} = 0 does not match the second derivative relationship. This is the standard JEE Advanced elimination technique. As a final consistency check, the resulting equation is second order with two constants eliminated, matching the requirement that order equals the number of arbitrary constants in the family.

Formation Of Differential Equations

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