Let y = y(x) be the solution of the differential equation $$x^{4}dy+(4x^{3}y+2\sin x)dx=0,x > 0,y\left(\frac{\pi}{2}\right)=0$$. Then $$\pi^{4}y\left(\frac{\pi}{3}\right)$$ is equal to :

We start with the given differential equation for x > 0: $$x^4\,dy + (4x^3y + 2\sin x)\,dx = 0.$$ Rewriting in the form of $$\frac{dy}{dx}$$ gives $$x^4\frac{dy}{dx} + 4x^3y + 2\sin x = 0.$$

Divide both sides by $$x^4$$ to obtain the standard linear form: $$\frac{dy}{dx} + \frac{4}{x}\,y = -\frac{2\sin x}{x^4} \quad -(1)$$

We use the integrating factor method. The integrating factor is given by $$\mu(x) = e^{\int P(x)\,dx},$$ where $$P(x) = \frac{4}{x}$$. Hence $$\mu(x) = e^{\int \frac{4}{x}\,dx} = e^{4\ln x} = x^4 \quad -(2)$$

Multiply equation (1) by $$\mu(x)=x^4$$. The left side becomes the derivative of the product $$x^4y$$, and the right side simplifies as follows: $$x^4\frac{dy}{dx} + 4x^3y = \frac{d}{dx}(x^4y),$$ $$x^4\cdot\Bigl(-\frac{2\sin x}{x^4}\Bigr) = -2\sin x.$$ Therefore we get $$\frac{d}{dx}(x^4y) = -2\sin x \quad -(3)$$

Integrate both sides of (3) with respect to $$x$$: $$\int \frac{d}{dx}(x^4y)\,dx = \int -2\sin x\,dx$$ gives $$x^4y = 2\cos x + C \quad -(4)$$

Use the initial condition $$y\bigl(\tfrac{\pi}{2}\bigr)=0$$ in (4): $$\Bigl(\tfrac{\pi}{2}\Bigr)^4\cdot 0 \;=\; 2\cos\!\Bigl(\tfrac{\pi}{2}\Bigr) + C \;\Longrightarrow\; 0 = 2\cdot 0 + C \;\Longrightarrow\; C = 0.$$ Thus the particular solution is $$y = \frac{2\cos x}{x^4}.$$

Finally, we compute $$\pi^4\,y\!\Bigl(\tfrac{\pi}{3}\Bigr) = \pi^4\cdot \frac{2\cos\!\bigl(\tfrac{\pi}{3}\bigr)}{\bigl(\tfrac{\pi}{3}\bigr)^4} = \pi^4\cdot \frac{2\cdot \tfrac12}{\tfrac{\pi^4}{3^4}} = 1\cdot \frac{3^4}{1} = 81.$$

Therefore, $$\pi^4\,y\!\Bigl(\tfrac{\pi}{3}\Bigr) = 81.$$ Option A is the correct answer.