Let `y=g(x)` be the solution of the differential equation `sinx((dy)/(dx))+y cos x=4x, If y(pi/2)=0`, then y(pi/6)` is equal to

Options

A. (a) `4/(9sqrt3)pi^(2)`
B. (b) `(-8)/(9sqrt3)pi^(2)`
C. (c) `-8/(9)pi^(2)`
D. (d) `-4/(9)pi^(2)`

Answer

(c)

Explanation

Wh have,
`sin x dy/dx +y cos x = 4x rArr dy/dx+ y cot x = 4 x cosec x`
This is a linear differential equation of form
`dy/dx+ Py=Q`
where `P=cot x,Q=4x cosec x`
Now, `IF=e^(int pdx)=e^(int cot xdx)=e^(logsinx)=sinx`
Solution of the differential equation is
`y cdot sin x = int 4x cosec x sin x dx +C`
`rArr ysin x = int 4xdx +C = 2x^(2)+C`
Put `x=pi/2,y=0,` we get
`C= -pi^(2)/2 rArr y sin x = 2x^(2) - pi^(2)/2`
Put `x =pi/6`
`therefore y(1/2)=2(pi^(2)/36)-pi^(2)/2`
`rArr y=pi^(2)/9-pi^(2) rArr y=-(8pi^(2))/9`
Alternate Method
We have , `sin x dy/dx+y cos x = 4x,` which can be written as
`d/dx(sin x cdoty) =4x`
On integrating both sides, we get
`int d/dx (sin x cdoty)cdot dx =int 4x cdot dx`
`rArr y cdot sin x =(4x^(2))/2 + C rArr y cdot sin x =2x^(2)+C`
Now, as y=0 when `x = pi/2`
`therefore C=-pi^(2)/2`
`rArr y cdot sin x = 2x^(2) -pi^(2)/2`
Now, putting `x=pi/6,` we get
`y(1/2)=2((pi^(2))/36)-pi^(2)/2 rArr y = pi^(2)/9-pi^(2)=-(8pi^(2))/9`