Solution of the equation `x^(2)y - x^(3)(dy)/(dx)=y^(4)cosx `, when y(0) =1 is

Options

A. `y^(3) = 3x^(3) sin x`
B. `x^(3) = 3y^(3) sin x`
C. ` x^(3) = y^(3) sinx`
D. `x^(3) = y^(3) cos x`

Answer

b

Explanation

Given that ,`x^(2)y - x^(3) (dy)/(dx) = y^(4) cos x`
i.e ` x^(3) (dy)/(dx) -x^(2) y = - y^(4) cos x`
On dividing `-y^(4)x^(3)` , we get
` -1/(y^(4)) (dy)/(dx) +1/(y^(3))* 1/x = 1/(x^(3))cos x`
Put ` 1/(y^(3)) = V rArr -1/(y^(4))(dy)/(dx) = 1/3 (dv)/(dx)`
` :. 1/3 (dv)/(dx) +1/xV = 1/(x^(3))`
` rArr (dv)/(dx) +3/x V = 3/(x^(3))cosx`
whch is linear differential equation V .
` :. IF = e^(int3/xdx) = e^(3 logx )=x^(3)`
So , the solution is ` x^(3) V = int x^(3) * 3/(x^(3)) cos x dx + C`
` = 3 sin x + C`
` rArr (x^(3))/(y^(3))= 3 sin x +C" " [ :' V = 1/(y^(3))]`
Putting x = 0 , y = 1, we get C = 0
Hence , the solution is `x^(3) = 3y^(3)` sin x