Let y=y(x) be the solution of the differential equation sinx(dy/dx) + ycosx = 4x, x∈(0,π). If y(π/2) = 0, then y(π/6) is equal to

Given differential equation: sinx(dy/dx) + ycosx = 4x where x∈(0,π)
∴ dy/dx + ycosx/sinx = 4x/sinx
dy/dx + cotxy = 4x/sinx
Comparing with dy/dx + P(x)y = Q(x)
∴ P(x) = cot(x), Q(x) = 4x/sinx
I.F = e∫P(x)dx = e∫cotxdx = e∫cotxdx = e^(log(|sinx|)) = sinx
∴ solution is given by yI.F = ∫Q(x)IFdx + c
sinx = ∫4x/sinx * sinx dx + c
= ∫4x dx + c
= 4x²/2 + c = 2x² + c
∴ ysinx = 2x² + c. (A)
given that y(π/2) = 0
i.e when x = π/2, y = 0
Equation becomes : 0 = 2(π/2)² + c
0 = 2 × π²/4 + c
∴ c = -π²/2
put this value in (A)
∴ ysinx = 2x² - π²/2
Next to find y(π/6) take x = π/6
∴ ysin(π/6) = 2(π/6)² - π²/2
y(1/2) = 2π²/36 - π²/2
y/2 = 2π²/36 - π²/2 × 9/9
y/2 = π²/18 - 9π²/18
y/2 = -8π²/18 = -4π²/9
y = -8π²/9
y = -8π²/9
∴ y(π/6) = -8π²/9
∴ y(π/6) = −π/2