The solution of the differential equation d y d x + y 2 sec ⁡ x = tan ⁡ x 2 y , where 0 ≤…

The solution of the differential equation dydx+y2sec⁡x=tan⁡x2y, where 0≤x<π2 and y0=1, is given by
  1. y2=1+xsec⁡x+tan⁡x
  2. y=1+xsec⁡x+tan⁡x
  3. y=1-xsec⁡x+tan⁡x
  4. y2=1-xsec⁡x+tan⁡x

Solution

dydx+y2sec⁡x=tan⁡x2y
2ydydx+y2sec⁡x=tan⁡x
Put y2=t ⇒2ydydx=dtdx
dtdx+tsec⁡x=tan⁡x
 I.F=e∫sec⁡xdx=eln(sec⁡x+tanx)=sec⁡x+tan⁡x
⇒tsec⁡x+tan⁡x=∫sec⁡x+tan⁡xtan⁡xdx
⇒tsec⁡x+tan⁡x=∫sec⁡xtan⁡xdx+∫tan2⁡xdx
⇒y2sec⁡x+tan⁡x=sec⁡x+tan⁡x-x+c
y0=1 ⇒c=0
⇒y2=1-xsec⁡x+tan⁡x

Asked in: JEE Main 2016 (10 Apr Online)

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