CAIE A-Level Mathematics A2 3.8.2 Differential Equations Questions

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.