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

Integrating the left-hand side:\n[ \int \frac{dy}{y^2} = -\frac{1}{y} + C_1 ]\n Integrating the right-hand side gives: [ \int \frac{1}{3\cos^2 2x} dx = \frac{1}{3} \tan 2x + C_2 ]\n Thus, combining these results gives: $-\frac{1}{y} = \frac{1}{3} \tan 2x + C$

Solving for $y$ provides: $y = -\frac{1}{\left(\frac{1}{3} \tan 2x + C\right)}$

Using the condition $y = 2$ when $x = -\frac{\pi}{8}$: [ 2 = -\frac{1}{\left(\frac{1}{3} \tan(-\frac{\pi}{4}) + C\right)} ] This implies: [ 2 = -\frac{1}{\left(-\frac{1}{3} + C\right)} ] Solving this for $C$ will yield the specific solution.

Returning from the previous expression, the final equation in the form $y = f(x)$ will be: $y = -\frac{1}{\frac{1}{3} \tan 2x + C}$, where $C$ has been determined from the initial condition.