Animated Solution for Mathematics - Differential Equations: Let $y=y(x)$ be the solution of the differential equation $2\cos x\frac{dy}{dx}=\sin 2x-4y\sin x,x\in (0,\frac{\pi }{2}).$ If $y(\frac{\pi }{3})=0$ then $y^{\prime }(\frac{\pi }{4})+y(\frac{\pi }{4})$ is equal to

Welcome, my dear student. Today, we are not just solving a problem; we are embarking on a journey through the elegant architecture of calculus. When you first look at the equation $2\cos x\frac{dy}{dx}=\sin 2x-4y\sin x$, it is natural to feel a momentary hesitation.

It looks cluttered and messy. But in the world of JEE Advanced, complexity is often just a mask for a very structured, very beautiful underlying order. Let us peel back that mask together.

Our first objective is to bring order to chaos. We want to transform this equation into the standard form of a Linear Differential Equation: $\frac{dy}{dx}+P(x)y=Q(x)$.

Starting with $2\cos x\frac{dy}{dx}=\sin 2x-4y\sin x$, we move the $y$ term to the left to obtain $2\cos x\frac{dy}{dx}+4y\sin x=\sin 2x$. To isolate $\frac{dy}{dx}$, we divide the entire equation by $2\cos x$:

We know the double angle identity $\sin 2x=2\sin x\cos x$. Substituting this into the right side, the $2\cos x$ terms cancel out, leaving us with $\sin x$. Our equation is now in its pristine standard form:

We have successfully identified $P(x)=2\tan x$ and $Q(x)=\sin x$. We have tamed the beast.

Now, we need the key that unlocks the solution: the Integrating Factor, or $IF$. The $IF$ is a mathematical construct designed to turn the left side of our equation into the derivative of a product, defined by $IF=e^{\int P(x)dx}$.

Substituting our $P(x)$:

We know that the integral of $\tan x$ is $\ln (\sec x)$. Thus, we have $e^{2\ln (\sec x)}$, which simplifies via the power rule for logarithms to $e^{\ln ({\sec }^{2}x)}$. Since the exponential and the natural logarithm are inverse functions, we arrive at the elegant result:

The general solution for a linear differential equation is given by $y\cdot (IF)=\int Q(x)\cdot (IF)dx+C$. Plugging in our values, we get:

Do not panic at the integral on the right. We can rewrite $\sin x{\sec }^{2}x$ as $\sin x\cdot \frac{1}{\cos x}\cdot \sec x$, which is simply $\tan x\sec x$. Since the integral of $\tan x\sec x$ is $\sec x$, our equation becomes:

Dividing by ${\sec }^{2}x$ (which is equivalent to multiplying by ${\cos }^{2}x$), we obtain our general solution:

We have a family of curves, but we need the specific one that satisfies the boundary condition $y(\frac{\pi }{3})=0$. Substituting $x=\frac{\pi }{3}$ and $y=0$:

Since $\cos (\frac{\pi }{3})=\frac{1}{2}$, we have $0=\frac{1}{2}+C(\frac{1}{4})$, which yields $C=-2$. Our particular solution is now fully defined:

Finally, the question asks for $y^{\prime }(\frac{\pi }{4})+y(\frac{\pi }{4})$. First, we find the derivative $y^{\prime }(x)$:

Evaluating these at $x=\frac{\pi }{4}$:

Adding these two results together, the $\frac{1}{\sqrt{2}}$ terms cancel out, and we are left with $2-1=1$. The final answer is 1.