CAIE A-Level Mathematics A2 3.8 Differential Equations Questions

State or imply equation of the form $\frac{\mathrm{d} x}{\mathrm{~d} t}=k x(300-x)$ and use $\frac{\mathrm{d} x}{\mathrm{~d} t}=0.2$ and x=1
M0 for verification.

Obtain $k=\frac{1}{1495}$ and rearrange to the given answer
$1495 \frac{\mathrm{~d} x}{\mathrm{~d} t}=x(300-x)$.

Separate variables correctly
$$\int(1+y) \mathrm{e}^{-3 y} \mathrm{~d} y=\int \frac{1}{1+\cos 2 \theta} \mathrm{~d} \theta$$

Marking guidance:

Allow $1 / \mathrm{e}^{3 y}$ and missing integral signs.

Integrate to obtain $p(1+y) \mathrm{e}^{-3 y}+\int q \mathrm{e}^{-3 y} \mathrm{~d} y$
Allow unless clear evidence that formula used has
a + sign.

Obtain $\frac{-1}{3}(1+y) \mathrm{e}^{-3 y}+\int \frac{1}{3} \mathrm{e}^{-3 y} \mathrm{~d} y$
Allow unsimplified.

Obtain $\frac{-1}{3}(1+y) \mathrm{e}^{-3 y}-\frac{1}{9} \mathrm{e}^{-3 y}(+A)$
Condone no constant of integration.

Use correct double angle formula to obtain $\int \frac{1}{2 \cos ^{2} \theta} \mathrm{~d} \theta$
Obtain $k \tan \theta[+B]$
Condone no constant of integration.

Use $y=0, \theta=\frac{\pi}{4}$ to evaluate a constant of integration in an expression of the form
$\alpha y \mathrm{e}^{-3 y}, \beta \mathrm{e}^{-3 y}$ and $\gamma \tan \theta$ only.

M1*

$$\frac{1}{2}=-\frac{1}{3}-\frac{1}{9}+C \quad\left(C=\frac{17}{18}\right)$$

Allow $\alpha y \mathrm{e}^{3 y}$ and $\beta \mathrm{e}^{3 y}$. Must have integrated LHS twice.

Use y=1
$$\frac{-(1+1)}{3 \mathrm{e}^{3}}-1\left(9 \mathrm{e}^{3}\right)=\frac{1}{2} \tan \theta-\frac{17}{18} .$$

Must have integrated LHS.

Obtain $\tan \theta=\frac{17}{9}-\frac{14}{9} \mathrm{e}^{-3}$
Or exact equivalent . Exact ISW.
$$\text { Allow } \theta=\tan ^{-1}\left(\frac{17}{9}-\frac{14}{9} \mathrm{e}^{-3}\right) \text {. }$$

If x instead of $\theta$ then withhold final A1.

Separate variables: $\int \frac{\sin \theta}{\cos ^{4} \theta} \mathrm{~d} \theta=\int \frac{(x+3)}{\left(x^{2}+9\right)} \mathrm{d} x$
Or $\int \frac{3 \sin \theta}{\cos ^{4} \theta} \mathrm{~d} \theta=\int \frac{3 x+9}{x^{2}+9} \mathrm{~d} x$.

Condone missing integral signs or missing d x or $\mathrm{d} \theta$, but not both.

Obtain $p \sec ^{3} \theta(+A)$
Correct form, p any constant but not 0 .

Use $\int \frac{3 x+9}{x^{2}+9} \mathrm{~d} x=\int\left(\frac{3 x}{x^{2}+9}+\frac{9}{x^{2}+9}\right) \mathrm{d} x$ and obtain $q \ln \left(x^{2}+9\right)$ or
$r \tan ^{-1} \frac{x}{3}(+C)$.

*M1

Might have one third of both sides.

Alt: substitute $x=3 \tan \phi$ to obtain $q \int 1+\tan \phi \mathrm{d} \phi$;
condone if have $\theta$ in place of $\phi$ in this method.

Obtain $q \ln \left(x^{2}+9\right)$ and $r \tan ^{-1} \frac{x}{3}(+C)$
Obtain $q(\phi \mp \ln (\cos \phi))$ OE.

Obtain $\sec ^{3} \theta=\frac{3}{2} \ln \left(x^{2}+9\right)+3 \tan ^{-1} \frac{x}{3}(+C)$ or equivalent
Or might see a third of both sides. Must have 2 different variables.

Use $\theta=\frac{1}{3} \pi, x=3$ in an equation including $p \sec ^{3} \theta, q \ln \left(x^{2}+9\right)$
and $r \tan ^{-1} \frac{x}{3}$ to evaluate the constant of integration
Or as limits in a definite integral.

Limits for $\phi$ are 0 and $\frac{1}{4} \pi$.

Obtain constant $=8-\frac{3}{2} \ln 18-\frac{3}{4} \pi$
OE, e.g. 1.308... to at least 3sf.

Obtain $\cos \theta=0.601$
Marking guidance:

Accept AWRT 0.601.