Animated Solution for Mathematics - Differential Equations: If $y=y(x),x\in (0,\frac{\pi }{2})$ be the solution curve of the differential equation $({\sin }^{2}2x)\frac{dy}{dx}+(8{\sin }^{2}2x+2\sin 4x)y=2e^{-4x}(2\sin 2x+\cos 2x)$, with $y(\frac{\pi }{4})=e^{-\pi }$, then $y(\frac{\pi }{6})$ is equal to :

Welcome, future engineers. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of trigonometric functions and exponential terms.

When you see an equation like

it is natural to feel a spike of anxiety. But I want you to take a deep breath. In the world of JEE Advanced, complexity is often just a mask. Our job is to peel back that mask and reveal the elegant structure underneath.

The Standardization

Every journey begins with a single, decisive step. In differential equations, that step is normalization. We cannot work with a coefficient attached to our $\frac{dy}{dx}$.

So, we divide the entire equation by ${\sin }^{2}2x$. By doing this, we transform our equation into the standard linear form:

Now, look at $P(x)=\frac{8{\sin }^{2}2x+2\sin 4x}{{\sin }^{2}2x}$. It looks messy, but remember your trigonometric identities! We know that $\sin 4x=2\sin 2x\cos 2x$.

Substituting this in, the expression simplifies beautifully. The ${\sin }^{2}2x$ terms cancel out, and we are left with $P(x)=8+4\cot 2x$. Suddenly, the chaos has order. We have tamed the coefficient.

The Magic Multiplier

Now we enter the heart of the method: the Integrating Factor (I.F.). This is the 'magic' that allows us to collapse the left side of our equation into a single derivative. The formula is $I.F.=e^{\int P(x)dx}$.

Integrating $8+4\cot 2x$ is straightforward. The integral of $8$ is $8x$, and the integral of $4\cot 2x$ is $2\ln |\sin 2x|$.

When we raise $e$ to the power of this sum, we use the laws of exponents to split it: $e^{8x}\cdot e^{2\ln (\sin 2x)}$. Using the logarithmic property $a\ln b=\ln b^a$, the second term becomes $e^{\ln ({\sin }^{2}2x)}$, which simplifies perfectly to ${\sin }^{2}2x$.

Our Integrating Factor is $e^{8x}{\sin }^{2}2x$. It is elegant, compact, and ready to work for us.

The Moment of Clarity

With the I.F. in hand, we multiply the entire differential equation by it. On the left side, we get the derivative of the product:

On the right side, the magic happens. When we multiply $Q(x)$ by our I.F., the ${\sin }^{2}2x$ in the denominator of $Q(x)$ cancels out with the ${\sin }^{2}2x$ in our I.F. We are left with:

Many students would immediately reach for integration by parts here. But stop! Look at the integrand: $4e^{4x}\sin 2x+2e^{4x}\cos 2x$.

Does this look familiar? It is the exact result of the product rule applied to $e^{4x}\sin 2x$. If you differentiate $e^{4x}\sin 2x$, you get exactly this expression. Recognizing this is the difference between a student who struggles and a student who masters the exam. The integral is simply $e^{4x}\sin 2x+C$.

The Final Victory

We are almost there. We have the general solution:

We use the initial condition $y(\frac{\pi }{4})=e^{-\pi }$ to find $C$. Substituting $x=\frac{\pi }{4}$, we find that $C=0$. This simplifies our function to:

Finally, we evaluate at $x=\frac{\pi }{6}$. Substituting this value, we get $y(\frac{\pi }{6})=\frac{e^{-2\pi /3}}{\sin (\pi /3)}$. Since $\sin (\frac{\pi }{3})=\frac{\sqrt{3}}{2}$, our final answer is:

Look at what you have achieved. You took a terrifying equation and, through systematic application of calculus, reduced it to a simple evaluation. This is the power of mathematics. Keep practicing, keep looking for the patterns, and never let the complexity intimidate you.