Differential Equations
Differential Equations
star_batch_jee_advanced_2025
Grade 12
Question:
Consider the differential equation $\cos^2 x \frac{dy}{dx} - (\tan 2x)y = \cos^4 x$; $|x| < \frac{\pi}{4}$ and $y\left(\frac{\pi}{6}\right) = \frac{3\sqrt{3}}{8}$ then:
Integrating factor of the differential equation is $\frac{\cos 2x}{1 + \cos 2x}$
Solution is $y = \frac{1}{2} \ln 2 \cos^2 x$
Solution is $y = \tan 2x \cos^2 x - \frac{3\sqrt{3}}{8}$
$y\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{8}$
Step-by-Step Solution
Key Concept: Linear first-order ODEs are solved via integrating factors to reduce to exact form.
For $\frac{dy}{dx} = \frac{\tan 2x}{\cos^2 x} = y\cos^2 x$, the integrating factor is $e^{\int\frac{2\sin 2x}{\cos^2 x}dx} = e^{\frac{\cos 2x}{1 + \cos 2x}}$. Solving yields $y = \frac{1}{2}\tan 2x \cos^2 x + c$. Using $y(\frac{\pi}{6}) = \frac{3\sqrt{3}}{8}$ gives $c = 0$, so $y = \frac{1}{2}\tan 2x\cos^2 x$.
Correct Answer: 1,2,4