Question

Solve the differential equation: $$ \frac{dy}{dx} = \cos^2(3x + y) $$

• A. \( 2\sqrt{3}(x + C) \)
• B. \( x + C \)
• C. \( \dfrac{x + C}{2\sqrt{3}} \)
• D. \( \dfrac{\sqrt{3}}{2}(x + C) \)

Hint:

For nonlinear ODEs involving trigonometric expressions in \( y \), try a substitution like \( z = ax + y \), and verify with a given function.

Answer 1

The Correct Option is A

Given: \[ \frac{dy}{dx} = \cos^2(3x + y) \Rightarrow \text{Use substitution } z = 3x + y \Rightarrow \frac{dz}{dx} = 3 + \frac{dy}{dx} \] Now: \[ \frac{dy}{dx} = \cos^2(z) \Rightarrow \frac{dz}{dx} = 3 + \cos^2(z) \] Separate: \[ \frac{dz}{3 + \cos^2(z)} = dx \Rightarrow \text{This is hard to integrate directly. Instead, solve the given using a known substitution.} \] Alternate approach (from the question’s suggestion):
We are told: \[ f(x) = \tan^{-1}\left(\frac{\sqrt{3}}{2} \tan(3x + y)\right) \] Then differentiate both sides: \[ \frac{df}{dx} = \frac{\sqrt{3}}{2} \cdot \frac{1}{1 + \left(\frac{\sqrt{3}}{2} \tan(3x + y)\right)^2} \cdot \frac{d}{dx}[\tan(3x + y)] \] \[ = \frac{\sqrt{3}}{2} \cdot \frac{1}{1 + \frac{3}{4}\tan^2(3x + y)} \cdot \sec^2(3x + y)\cdot (3 + \frac{dy}{dx}) \] Solve this algebraically yields the ODE solution: \[ f(x) = 2\sqrt{3}(x + C) \]