Question

12. The equation of trajectory of a particle is
x=v3y - 5y”. The particle is at origin at t = 0. Its
acceleration is a =-101. What is its initial speed?​

Answer 1

Multiplying by we have ,, ^ ^ ’

A cos .r + ^ sin x. (12)

Two differentiations give

' 4^ 2/y) — —A sin .r *f- B cos .r, (13)

^ 4yy' -|- 4?/) ~ - A cos X — sin x. ( 14)

Thus, the effect of multiplying the original equation by e-' was an

equation (12) whose right member reproduced itself, apart from sign,

after two differentiations. Adding equations (12) and (14), we obtain

the required differential equation

. //" 4/y' f 5y - 0. (15)

Answer 2

Explanation:

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