Animated Solution for Mathematics - Differential Equations: If $\cos x\frac{dy}{dx}-y\sin x=6x,(0<x<\frac{\pi }{2})$ and $y(\frac{\pi }{3})=0$, then $y(\frac{\pi }{6})$ is equal to :-

Welcome, student! Today, we are going to unravel a differential equation that might look intimidating at first glance, but beneath its surface lies a beautiful, logical structure.

We are given the equation $\cos x\frac{dy}{dx}-y\sin x=6x$ with the initial condition $y(\frac{\pi }{3})=0$. Our mission is to find the value of $y$ at $x=\frac{\pi }{6}$.

The Art of Standardization

In the world of differential equations, the first step is often the most critical. We are looking for the standard linear form: $\frac{dy}{dx}+P(x)y=Q(x)$.

Currently, our equation is $\cos x\frac{dy}{dx}-y\sin x=6x$. To get $\frac{dy}{dx}$ by itself, we divide the entire equation by $\cos x$.

This gives us:

Since $\frac{\sin x}{\cos x}=\tan x$ and $\frac{1}{\cos x}=\sec x$, our equation transforms into the elegant form:

Now, we can clearly identify our functions: $P(x)=-\tan x$ and $Q(x)=6x\sec x$.

The Magic of the Integrating Factor

Now that we have our $P(x)$, we need the Integrating Factor (IF). The formula is $IF=e^{\int P(x)dx}$.

Substituting our $P(x)$, we get $IF=e^{\int -\tan xdx}$. We know that the integral of $\tan x$ is $\ln (\sec x)$, so the integral of $-\tan x$ is $-\ln (\sec x)$, which is equivalent to $\ln (\cos x)$.

Thus:

Because the exponential and natural logarithm are inverse functions, they cancel out perfectly, leaving us with $IF=\cos x$. This is the magic key that will simplify our equation.

The General Solution

With our IF in hand, we multiply our standardized equation by $\cos x$. The left side of our equation becomes the derivative of the product $y\cdot IF$.

Specifically:

Since $\sec x\cdot \cos x=1$, the right side simplifies beautifully to $6x$. Now, we integrate both sides:

This gives us the general solution:

The Particular Solution and Final Reveal

We are almost there! We use the initial condition $y(\frac{\pi }{3})=0$ to find $C$.

Substituting $x=\frac{\pi }{3}$ and $y=0$ into $y\cos x=3x^2+C$:

Thus, $C=-\frac{{\pi }^2}{3}$. Our particular solution is $y\cos x=3x^2-\frac{{\pi }^2}{3}$.

Finally, to find $y(\frac{\pi }{6})$, we substitute $x=\frac{\pi }{6}$:

Since $\cos (\frac{\pi }{6})=\frac{\sqrt{3}}{2}$, we have:

Solving for $y(\frac{\pi }{6})$, we get:

You have successfully navigated the complexity of this differential equation! The final answer is $-\frac{{\pi }^2}{2\sqrt{3}}$.