The motion of a particle is described by the differential equation \(\frac{{{\mathrm{d}}^{2}}x}{\mat

Question

The motion of a particle is described by the differential equation $\frac{{{\mathrm{d}}^{2}}x}{\mathrm{d}{{t}^{2}}}-4\frac{\mathrm{d}x}{\mathrm{d}t}+4x={{\mathrm{e}}^{2t}}$, where $x$ is the displacement of the particle in metres along the $x$- axis with respect to origin $O$ at time $t$ seconds. By using the substitution $x=u{{\mathrm{e}}^{2t}}$, show that this differential equation may be reduced to the form $\frac{{{\mathrm{d}}^{2}}u}{\mathrm{d}{{t}^{2}}}=a$, where $a$ is a constant to be determined. Hence, find the displacement of the particle at time $t$ if the particle moves off from the initial point $x=1$ with a velocity of $1$m/s.[7] After some time, the particle is returned to its initial point and set into a new motion along the $x$- axis. The motion of the particle is described by the differential equation $\frac{\mathrm{d}v}{\mathrm{d}t}=2{{\mathrm{e}}^{-v}}$, where $v$ is the velocity of the particle in metres per second at time $t$ seconds after it starts on this new motion. If the particle is at rest just before it begins this new motion, find an expression for $t$ in terms of $v$. Show that $2x=3+{{\mathrm{e}}^{v}}(v-1)$, where $x$ is the horizontal displacement of the particle in metres.[6]

Accepted Answer

$\frac{{{	ext{d}}^{2}}u}{	ext{d}{{t}^{2}}}=1$; $x={{	ext{e}}^{2t}}\left( \frac{{{t}^{2}}}{2}-t+1 ight)=\frac{{{e}^{2t}}}{2}\left( {{t}^{2}}-2t+2 ight)$;${{	ext{e}}^{v}}=2t+k$, where $k$ is an arbitrary constant.

2020 NYJC CT2 P1 Q10

2020 NYJC CT2 P1 Q10

Video Solution: