Let y=y(x) be a solution of the differential equation (x cos x) d y+(x y sin x+y cos x-1) d x=0,0 < x < π/2. If π/3 y(π/3)=√3, then |π/6 y^(′ ′)(π/6)+2 y′(π/6)| is equal to ____________.

Solution

Maths sir · 4 steps

Spot that xy''+2y' is the second derivative of xy, so solve for xy and differentiate it twice.

1. Linear form. Divide by x cos x dx to get dy/dx + (tan x + 1/x)y = 1/(x cos x). The integrating factor is e^(∫(tan x + 1/x) dx) = x/(cos x) = x sec x.
2. Integrate. Multiplying through gives d/dx(xy sec x) = sec² x, so xy sec x = tan x + C. At x = π/3, sec x = 2 and xy = √3, so 2√3 = √3 + C and C = √3.

   xy = sin x + √3 cos x

3. Differentiate twice. Once: y + xy' = cos x - √3 sin x. Again: xy'' + 2y' = -sin x - √3 cos x. That left side is exactly the expression in the question.
4. So. At x = π/6: -1/2 - √3 × √3/2 = -1/2 - 3/2 = -2. Its absolute value is 2.

The common slip: -2. Dropping the modulus leaves -2; the question asks for the absolute value, which is 2.

The idea to keep. When a question asks for a combination like xy''+2y', check whether it is the derivative of something simple such as (xy)''. Then you never need y'' on its own.

Answer from the NTA final answer key. Solution by PrepWiser.