Differential Equations

The motion of a particle is described by the differential equation $\frac{{{\text{d}}^{2}}x}{\text{d}{{t}^{2}}}-4\frac{\text{d}x}{\text{d}t}+4x={{\text{e}}^{2t}}$, 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 $x=u{{\text{e}}^{2t}}$, show that this differential equation may be reduced to the form $\frac{{{\text{d}}^{2}}u}{\text{d}{{t}^{2}}}=a$, 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 $\frac{\text{d}v}{\text{d}t}=2{{\text{e}}^{-v}}$, 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 $2x=3+{{\text{e}}^{v}}(v-1)$, where $x$ is the horizontal displacement of the particle in metres.

[6]

In established forest fires, the proportion of the total area of the forest which has been destroyed is denoted by $x$, and the rate of change of $x$ with respect to time, $t$ hours, is called the destruction rate. Investigations show that the destruction rate is directly proportional to the product of the proportion destroyed and the proportion not destroyed. A particular fire is initially noticed when one half of the forest is destroyed, and it is found that the destruction rate at this time is such that, if it remained constant thereafter, the forest would be destroyed completely in a further 24 hours.

Show that $12\frac{\text{d}x}{\text{d}t}=x(1-x)$ .

[3]

Deduce that approximately 73% of the forest is destroyed 12 hours after it is first noticed.

[6]

A tank has a capacity of 100 litres. Initially, the tank contains 10 litres of water thoroughly mixed with 300 grams of salt. Salt water with a concentration of 5 g/litre is poured into the tank at a constant rate of 2 litres per minute, while the mixture flows out at a constant rate of 1 litre per minute.
Let $S$ denote the amount of dissolved salt in the tank (in grams) at time $t$ minutes after salt water is poured into the tank.

A spherical tank with negligible thickness and internal radius $a$cm contains water. At time $t$s, the water surface is at a height $x$cm above the lowest point of the tank and the volume of water in the tank, $V$cm$^{3}$, is given by $V=\frac{1}{3}\pi {{x}^{2}}\left( 3a-x \right)$. Water flows from the tank, through an outlet at its lowest point, at a rate $\pi k\sqrt{x}$cm$^{3}$ s$^{-1}$, where $k$ is a positive constant.