Reducing Differential Equations (Edexcel A Level Further Maths: Further Pure 1): Exam Questions

The vertical height, $h$ m, above horizontal ground, of a passenger on a fairground ride, $t$ seconds after the ride starts, where $t⩽5$, is modelled by the differential equation

$t^2\frac{{\text{d}}^{2}h}{\text{d}{t}^2}-2t\frac{\text{d}h}{\text{d}t}+2h=t^3\text{(I)}$

Given that $t={\text{e}}^x$, show that

(i) $t\frac{\text{d}h}{\text{d}t}=\frac{\text{d}h}{\text{d}x}$

(ii) $t^2\frac{{\text{d}}^{2}h}{\text{d}{t}^2}=\frac{{\text{d}}^{2}h}{\text{d}{x}^2}-\frac{\text{d}h}{\text{d}x}$

Hence show that the transformation $t={\text{e}}^x$ transforms equation (I) into the equation

$\frac{{\text{d}}^{2}h}{\text{d}{x}^2}-3\frac{\text{d}h}{\text{d}x}+2h={\text{e}}^{3x}$

Hence show that

$h=At+B{t}^2+\frac{1}{2}{t}^3$

where $A$ and $B$ are constants.

Given that when $t=1$, $h=2.5$ and when $t=2$, $\frac{\text{d}h}{\text{d}t}=-1$

Determine the height of the passenger above the ground 5 seconds after the start of the ride.

A particle $P$ moves along a straight line.

At time $t$ minutes, the displacement, $x$ metres, of $P$ from a fixed point $O$ on the line is modelled by the differential equation

$t^2\frac{{\text{d}}^{2}x}{\text{d}{t}^2}-2t\frac{\text{d}x}{\text{d}t}+2x+16t^{2}x=4t^3\text{sin}2t\text{(I)}$

Show that the transformation $x=ty$ transforms equation (I) into the equation

$\frac{{\text{d}}^{2}y}{\text{d}{t}^2}+16y=4\text{sin}2t$

Hence find a general solution for the displacement of $P$ from $O$ at time $t$ minutes.

A community is concerned about the rising level of pollutant in its local pond and applies a chemical treatment to stop the increase of pollutant.

The concentration, $x$ parts per million (ppm), of the pollutant in the pond water $t$ days after the chemical treatment was applied, is modelled by the differential equation

$\frac{\text{d}x}{\text{d}t}=\frac{3+\text{cosh}t}{3x^2\text{cosh}t}-\frac{1}{3}x\text{tanh}t\text{(I)}$

When the chemical treatment was applied the concentration of pollutant was 3 ppm.

Use the iteration formula

${\left(\frac{\text{d}y}{\text{d}x}\right)}_n\approx \frac{\left(y_{n+1}-y_n\right)}{h}$

once to estimate the concentration of the pollutant in the pond water 6 hours after the chemical treatment was applied.

Show that the transformation $u=x^3$ transforms the differential equation (I) into the differential equation

$\frac{\text{d}u}{\text{d}t}+u\text{tanh}t=1+\frac{3}{\text{cosh}t}\text{(II)}$

Determine the general solution of equation (II)

Hence find an equation for the concentration of pollutant in the pond water $t$ days after the chemical treatment was applied.

Find the percentage error of the estimate found in part (a) compared to the value predicted by the model, stating if it is an overestimate or an underestimate.

The concentration of a drug in the bloodstream of a patient, $t$ hours after the drug has been administered, where $t\le 6$, is modelled by the differential equation

$t^2\frac{{\text{d}}^{2}C}{\text{d}{t}^2}-5t\frac{\text{d}C}{\text{d}t}+8C=t^3\text{(I)}$

where $C$ is measured in micrograms per litre.

Show that the transformation $t={\text{e}}^x$ transforms equation (I) into the equation

$\frac{{\text{d}}^{2}C}{\text{d}{x}^2}-6\frac{\text{d}C}{\text{d}x}+8C={\text{e}}^{3x}\text{(II)}$

Hence find the general solution for the concentration $C$ at time $t$ hours.

Given that when $t=6$, $C=0$ and $\frac{\text{d}C}{\text{d}t}=-36$

find the maximum concentration of the drug in the bloodstream of the patient.

A vibrating spring, fixed at one end, has an external force acting on it such that the centre of the spring moves in a straight line. At time $t$ seconds, $t\ge 0$, the displacement of the centre $C$ of the spring from a fixed point $O$ is $x$ micrometres.

The displacement of $C$ from $O$ is modelled by the differential equation

$t^2\frac{{\text{d}}^{2}x}{\text{d}{t}^2}-2t\frac{\text{d}x}{\text{d}t}+(2+t^2)x=t^4\text{(I)}$

Show that the transformation $x=tv$ transforms equation (I) into the equation

$\frac{{\text{d}}^{2}v}{\text{d}{t}^2}+v=t\text{(II)}$

Hence find the general equation for the displacement of $C$ from $O$ at time $t$ seconds.

(i) State what happens to the displacement of $C$ from $O$ as $t$ becomes large.

(ii) Comment on the model with reference to this long term behaviour.

The motion of a particle $P$ along the $x$-axis is modelled by the differential equation

$t^2\frac{{\text{d}}^{2}x}{\text{d}{t}^2}-t(3t+2)\frac{\text{d}x}{\text{d}t}+(2t^2+3t+2)x=12t^3{\text{e}}^{3t}\left(\text{I}\right)$

where $P$ has displacement $x$ metres from the origin $O$ at time $t$ minutes, $t>0$.

Show that the transformation $x=tu$ transforms the differential equation ($\mathrm{I}$) into the differential equation

$\frac{{\text{d}}^{2}u}{\text{d}{t}^2}-3\frac{\text{d}u}{\text{d}t}+2u=12{\text{e}}^{3t}$

Given that $P$ is at $O$ when $t=\ln 2$ and when $t=\ln 3$, determine the particular solution of the differential equation (I).