Question

he motion of a set of particles moving along the x-axis is governed by the differential equation; dt/dx t^3-x^3,where x(t) denotes the position at time t of the particle. Show that the acceleration of a particle is given by d^2x/dt^2=3t^2-3t^33x^2 +3x^5 .

Answer 1

\[d ^{2}x/dt ^{2}=3t ^{2}-3t ^{2}x ^{2}+3x ^{5}\] here is a better notation of equation as stated above

Answer 2

Alrighty:
\[ \frac{dx}{dt} = t^3 - x^3\]
\[ \frac{d^2x}{dt^2} = \frac{d}{dt} \left(\frac{dx}{dt}\right) = \frac{d}{dt}\left(t^3 - x^3\right) = 3t^2 - 3x^2\cdot \frac{dx}{dt} = 3t^2 - 3x^2\left(t^3-x^3\right) = 3t^2 - 3x^2t^3 + 3x^5\]

Answer 3

It was cut off, but the last part of that is your answer...