If y = y x is the solution of the differential equation d y d x + tan x y = sin x , 0 ≤ x ≤…

If y=yx is the solution of the differential equation dydx+tanxy=sinx,0xπ3, with y0=0, then yπ4 is equal to

  1. 14loge2
  2. 122loge2

  3. loge2
  4. 12loge2

Solution

dydx+tanxy=sinx; 0xπ3

I.F.=etanxdx=elnsecx=secx

ysecx=tanxdx+C

ysecx=lnsecx+C

x=0, y=0   C=0

ysecx=lnsecx

y=cosx·lnsecx

yx=π4=12·ln2

yx=π4=122loge2

Asked in: JEE Main 2021 (16 Mar Shift 2)

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