If y = a cos 2x + b sin 2x, then

Correct Answer is : d2ydx2+4y=0{{{d²y}}{{d{x²}} + 4y = 0dx2d2y+4y=0

Correct Answer is : d2ydx2+4y=0{{{d²y}}{{d{x²}} + 4y = 0dx2d2y+4y=0

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UPSC NDA I 2019 Mathematics Solved Paper

Question : 83 of 120

Marks: +1, -0

y = a cos 2x + b sin 2x

, then

Solution:

Concept:

The problem involves finding the second-order differential equation satisfied by a given trigonometric function.

Explanation:

Given

y = a \cos 2x + b \sin 2x

Differentiate with respect to

x

\frac{dy}{dx} = -2a \sin 2x + 2b \cos 2x

Differentiate again:

\frac{d^2y}{dx^2} = -4a \cos 2x - 4b \sin 2x

Factor out

-4

\frac{d^2y}{dx^2} = -4 (a \cos 2x + b \sin 2x) = -4y

Thus, the differential equation becomes

\frac{d^2y}{dx^2} + 4y = 0

This matches option D.

Answer:

\frac{{{d^2}y}}{{d{x^2}}} + 4y = 0

(Option D)

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