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 7 - 2018 - Paper 9

Next, we integrate both sides:

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

The left side integrates to:

$-\frac{1}{y}$

and for the right side, we utilize the identity $\sec^2(a) = \frac{1}{\cos^2(a)}$:

$\int \frac{1}{3}\sec^2(2x) dx = \frac{1}{3} \cdot \frac{1}{2} \tan(2x) + C = \frac{1}{6} \tan(2x) + C$

Thus, we have:

$-\frac{1}{y} = \frac{1}{6} \tan(2x) + C$ or rearranging gives:

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

Now, we apply the initial condition $y = 2$ when $x = -\frac{\pi}{8}$:

Substituting, we find:

$2 = -\frac{1}{\frac{1}{6} \tan\left(2(-\frac{\pi}{8})\right) + C}$

Calculating $tan$:

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

Solving for $C$ leads to $C = -\frac{1}{2}$.

Replacing $C$ back, the solution is:

$y = -\frac{1}{\frac{1}{6} \tan(2x) - \frac{1}{2}}$ which simplifies to:

$y = \frac{6}{3 \tan(2x) - 1}$ This gives us our final answer in the form $y = f(x)$.