Let y=y (x ) be the solution of the differential equation x^4 dy+(4x^3y+2 sinx) dx=0 ,x0 ,y ( 2 )=0. Then ^4y ( 3 ) is equal to:

Question

Let y=y (x ) be the solution of the differential equation x^4 dy+(4x^3y+2 sinx) dx=0 ,x>0 ,y ( 2 )=0. Then ^4y ( 3 ) is equal to:

Accepted Answer

Correct answer: D. (d) x^4 dy+4x^3 ydx+2 sinx dx=0 ⇒d (x^4y )+2 sinx dx=0. Integrating: x^4y-2 cosx=C. y ( /2 )=0 : ( /2 )^4 0-2 cos ( /2 )=0 ⇒C=0. x^4y=2 cosx. At x= /3 : ( /3 )^4y ( /3 )=2 cos ( /3 )=1. ^4y ( /3 )=81 ( /3 )^4y ( /3 ) 1 ( /3 )^4 ^481 … Directly: y ( /3 )= 81 ^4, so ^4y ( /3 )=81.

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

Solution

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