Given that $y = 2$ when $x = -\frac{\pi}{8}$ solve the differential equation $$\frac{dy}{dx} = \frac{y^2}{3\cos^2 2x}$$ $-\frac{1}{2} < x < \frac{1}{2}$ giving your answer in the form $y = f(x)$. - Edexcel - A-Level Maths Pure - Question 8 - 2018 - Paper 9

Next, we integrate both sides:

$\int \frac{1}{y^2} dy = \int \frac{1}{3\cos^2 2x} dx$

The left side integrates to:

$-\frac{1}{y} + C_1$

And the right side can be solved using the identity $\sec^2 u = \frac{d}{du} \tan u$:

$\int \sec^2 2x dx = \frac{1}{2} \tan 2x + C_2$.

Thus, we have:

$-\frac{1}{y} = \frac{1}{6} \tan 2x + C$.

Now we solve for $y$:

$y = -\frac{1}{\frac{1}{6} \tan 2x + C}$.

Given $y = 2$ when $x = -\frac{\pi}{8}$, we plug in these values:

$2 = -\frac{1}{\frac{1}{6} \tan(-\frac{\pi}{4}) + C}$

Knowing $\tan(-\frac{\pi}{4}) = -1$, we can solve for $C$:

$2 = -\frac{1}{-\frac{1}{6} + C} \Rightarrow -\frac{1}{2} = -\frac{1}{6} + C \Rightarrow C = \frac{1}{3}.$

Substituting back into the equation for $y$ gives:

$y = -\frac{1}{\frac{1}{6} \tan 2x + \frac{1}{3}} = -\frac{6}{\tan 2x + 2}.$

Thus, the solution in the required form is:

$y = f(x) = -\frac{6}{\tan 2x + 2}.$