QJEEM20Ma7b85: JEE Main 2020 Mathematics - Class 12 JEEM20Ma7b85 Solution: Let y = y(x) be the solution of the differential equation cosx…

Question

Verified Solution • 2025–26 Session moderate

Let y = y(x) be the solution of the differential equation

cosx

\frac{d y}{d x}

+ 2ysinx = sin2x, x

\in

(0, \frac{π}{2})

.

If y

(\frac{π}{3})

= 0, then y

(\frac{π}{4})

is equal to :

Accepted Answer

cosx

\frac{d y}{d x}

+ 2ysinx = sin2x

\Rightarrow

\frac{d y}{d x} + 2 y \tan x = 2 \sin x

I.F =

e^{\int 2 \tan x d x}

= sec² x

y.sec² x =

\int 2 \sin x \sec^{2} x d x

\Rightarrow

ysec²x =

\int 2 \tan x \sec x d x

\Rightarrow

ysec²x = 2secx + c

Given at x =

\frac{π}{3}

, y = 0

\Rightarrow

0 =

2 \sec \frac{π}{3} + c

\Rightarrow

c = -4

ysec²x = 2secx - 4

Here put x =

\frac{π}{4}

y.2 =

2 \sqrt{2}

- 4

\Rightarrow

y =

\sqrt{2} - 2

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