Let y=y(x) be the solution of the differential equation x^4 ~d y+ (4 x^3 y+2 sin x ) d x=0, x>0, y ( π 2 )=0. Then π^4 y ( π 3 ) is equal to :

$x^4\,dy + 4x^3 y\,dx + 2\sin x\,dx = 0 \Rightarrow d(x^4 y) + 2\sin x\,dx = 0$x4dy+4x3ydx+2sinxdx=0⇒d(x4y)+2sinxdx=0.

Integrating: $x^4 y - 2\cos x = C$x4y−2cosx=C.

$y(\pi/2) = 0$y(π/2)=0: $(\pi/2)^4 \cdot 0 - 2\cos(\pi/2) = 0 \Rightarrow C = 0$(π/2)4⋅0−2cos(π/2)=0⇒C=0.

$x^4 y = 2\cos x$x4y=2cosx.

At $x = \pi/3$x=π/3: $(\pi/3)^4 y(\pi/3) = 2\cos(\pi/3) = 1$(π/3)4y(π/3)=2cos(π/3)=1.

$\pi^4 y(\pi/3) = 81 \cdot (\pi/3)^4 y(\pi/3) \cdot \frac{1}{(\pi/3)^4} \cdot \frac{\pi^4}{81}$π4y(π/3)=81⋅(π/3)4y(π/3)⋅(π/3)41​⋅81π4​... Directly: $y(\pi/3) = \frac{81}{\pi^4}$y(π/3)=π481​, so $\pi^4 y(\pi/3) = 81$π4y(π/3)=81.