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Step-by-Step Solution
Step 1: Rewrite the differential equation in standard form
Given:
$ \cos x \frac{dy}{dx} - y \sin x = 6x $
First, divide through by $\cos x$ (noting that $0 < x < \frac{\pi}{2}$ implies $\cos x \neq 0$):
$ \frac{dy}{dx} - y \tan x = 6x \sec x. $
Step 2: Identify the integrating factor
A standard linear differential equation has the form:
$ \frac{dy}{dx} + P(x)\,y = Q(x). $
Here, $P(x) = -\tan x $ and $Q(x) = 6x \sec x.$
The integrating factor (IF) is given by:
$ \text{IF} = e^{\int P(x)\,dx} = e^{\int -\tan x \, dx}. $
We know:
$ \int \tan x \, dx = -\ln|\cos x| \quad \bigl(\text{equivalently, } \int -\tan x \, dx = \ln|\cos x|\bigr). $
Since $0 < x < \frac{\pi}{2}$, $\cos x$ is positive. Hence:
$ \text{IF} = e^{\ln(\cos x)} = \cos x. $
Step 3: Multiply the differential equation by the integrating factor
Multiply both sides of
$ \frac{dy}{dx} - y \tan x = 6x \sec x $
by $\cos x$:
$ \cos x \frac{dy}{dx} - y \sin x = 6x \sec x \cos x. $
But $ \sec x \cos x = 1 $, so:
$ \cos x \frac{dy}{dx} - y \sin x = 6x. $
Notice this is the original equation, and in integrating-factor form it becomes:
$ \frac{d}{dx}\bigl[y(\cos x)\bigr] = 6x.
$
Step 4: Integrate to find the general solution
Integrate both sides with respect to $x$:
$ y(\cos x) = \int 6x \, dx = 3x^2 + C, $
where $C$ is the constant of integration.
Step 5: Use the given condition to find the constant $C$
We are given $y\bigl(\tfrac{\pi}{3}\bigr) = 0$. Hence:
$ 0 \cdot \bigl(\cos(\tfrac{\pi}{3})\bigr) = 3\bigl(\tfrac{\pi}{3}\bigr)^{2} + C. $
Since $\cos\bigl(\tfrac{\pi}{3}\bigr) = \tfrac{1}{2}$, the left side is $y \cdot \tfrac{1}{2}$, but $y(\tfrac{\pi}{3})=0$ leads that product to 0. So:
$ 0 = 3\left(\frac{\pi^2}{9}\right) + C. $
Which simplifies to:
$ 0 = \frac{3\pi^2}{9} + C \quad \Rightarrow \quad C = - \frac{\pi^2}{3}. $
Step 6: Write the specific solution
Substitute $C$ back into $y(\cos x)$:
$ y \cos x = 3x^2 - \frac{\pi^2}{3}. $
Step 7: Find $y\bigl(\tfrac{\pi}{6}\bigr)$
For $x = \tfrac{\pi}{6}$:
$ y(\tfrac{\pi}{6}) \cos\bigl(\tfrac{\pi}{6}\bigr) = 3\bigl(\tfrac{\pi}{6}\bigr)^2 - \frac{\pi^2}{3}. $
We have $\cos\bigl(\tfrac{\pi}{6}\bigr) = \frac{\sqrt{3}}{2}$. Also, $(\tfrac{\pi}{6})^2 = \frac{\pi^2}{36}$. Thus:
$ y(\tfrac{\pi}{6}) \cdot \frac{\sqrt{3}}{2} = 3 \cdot \frac{\pi^2}{36} - \frac{\pi^2}{3}. $
Simplify the right-hand side:
$ 3 \cdot \frac{\pi^2}{36} = \frac{\pi^2}{12}, \quad \text{and} \quad - \frac{\pi^2}{3} = - \frac{4\pi^2}{12}. $
So:
$ \frac{\pi^2}{12} - \frac{4\pi^2}{12} = - \frac{3\pi^2}{12} = - \frac{\pi^2}{4}. $
Hence:
$ y(\tfrac{\pi}{6}) \cdot \frac{\sqrt{3}}{2} = - \frac{\pi^2}{4}. $
Solve for $ y(\tfrac{\pi}{6}) $:
$ y(\tfrac{\pi}{6}) = \left(- \frac{\pi^2}{4}\right) \Big/ \left(\frac{\sqrt{3}}{2}\right)
= - \frac{\pi^2}{4} \cdot \frac{2}{\sqrt{3}}
= - \frac{\pi^2}{2\sqrt{3}}. $
Step 8: State the final answer
Therefore,
$ \displaystyle y\bigl(\tfrac{\pi}{6}\bigr) = - \frac{\pi^2}{2\sqrt{3}}. $
