[" (iii) "x=(a+bt)e^(-nt)" ₹6% "6" hen "a" ,"],[(d^(2)x)/(dt^(2))+2n(dx)/(dt)+n^(2)x=0]

Q.if x=(a+bt)e^(-nt), show that,((d^(2)x)/(dt^(2)))+2n((dx)/(dt))+n^(2)=0

IF x=(a+bt)e^(-nt) , show that , (d^2x)/(dt^2)+2ndx/dt+n^2x=0 .

If x^2=t^2+! then find (d^2x)/(dt^2)

Show that , the solution x = e^(-kt) (a cos nt + b sin nt ), for all a and b , always satisfies the differenital equation (d^(2)x)/(dt^(2)) + 2k (dx)/(dt) + (k^(2) + n^(2)) x = 0

If x=a cos nt-b sin nt, then (d^(2)x)/(dt^(2)) is n^(2)x(b)-n^(2)x(c)-nx(d)nx

Order and degree of the differential equation (d^(2)x)/(dt^(2))+((dx)/(dt))^(2)+7=0 respectively are

If x=a cos nt-b sin nt, then (d^(2)x)/(dt^(2)) is n^(2)x( b) -n^(2)x( c) nx( d )nx

The equation of a simple harmonic motion of a particle is (d^(2)x)/(dt^(2)) + 0.2 (dx)/(dt) + 36x = 0 . Its time period is approximately

If x=a cos nt-b sin nt , then (d^2x)/(dt^2)=