QJEEM16Ma6dca: JEE Main 2016 Mathematics - Class 12 JEEM16Ma6dca Solution: The solution of the differential equation where 0 x < , and y…

Question

Verified Solution • 2025–26 Session moderate

The solution of the differential equation

{{dy} \over {dx}}\, + \,{y \over 2}\,\sec x = {{\tan x} \over {2y}},\,\,

where 0

\le

x <

{\pi \over 2}

, and y (0) = 1, is given by :

Accepted Answer

Given,

{{dy} \over {dx}} + {y \over 2}\sec x = {{\tan x} \over {2y}}

\Rightarrow

2y{{dy} \over {dx}} + {y^2}\sec x = \tan x

Now, let

y²

=

t

\Rightarrow

2y

{{dy} \over {dx}} = {{dt} \over {dx}}

\therefore

New equation,

{{dt} \over {dx}} + t\sec x = \tan x

\therefore

I.F

=

{e^{\int {\sec xdx} }}

=

{e^{\ln \left( {\sec x + \tan x} \right)}}

=

sec x + tan x

\therefore

Solution is,

t(sec x + tan x)

=

\int {\tan x}

(sec x + tan x) dx

\Rightarrow

t(sec x + tan x)

=

sec x + tan x

-

x + c

\Rightarrow

t

=

1

-

{x \over {\sec x + \tan x}} + c

\Rightarrow

y²

=

1

-

{x \over {\sec x + \tan x}} + c

Given,

y(0)

=

1

\therefore

1

=

1

-

0 + c

\Rightarrow

c

=

0

\therefore

y²

=

1

-

{x \over {\sec x + \tan x}}

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