QJEEM18Ma3936: JEE Main 2018 Mathematics - Class 12 JEEM18Ma3936 Solution: Let y = y(x) be the solution of the differential equation , .…

Question

Verified Solution • 2025–26 Session moderate

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

\sin x{{dy} \over {dx}} + y\cos x = 4x

,

x \in \left( {0,\pi } \right)

.

If

y\left( {{\pi \over 2}} \right) = 0

, then

y\left( {{\pi \over 6}} \right)

is equal to :

Accepted Answer

Step-by-Step Verified Solution

Given,

sin x

{{dy} \over {dx}} + y\cos y = 4x

\Rightarrow \,\,\,\,{{dy} \over {dx}}\,

+ y cot x = 4x cosec x

This is a linear differential equation of form,

{{dy} \over {dx}}\,

+ py = Q

Where p = cot x and Q = 4x cosec x

So, Integrating factor (I. F)

=

{e^{\int {pdx} }}

=

{e^{\int {\cot dx} }}

=

{e^{\ln \left| {\sin \,x} \right|}}

\left|="" {\sin="" x}="" \right|

=""
= sin x as

x \in \left( {0,\pi } \right)

Solution of the differential equation is

y sin x =

\int {}

4x cosecx sinx dx + c

\Rightarrow \,\,\,\,

y sinx =

\int {4x\,dx\, + c}

\Rightarrow \,\,\,\,

y sinx = 4.

{{{x^2}} \over 2} + c

\Rightarrow \,\,\,\,

y sinx = 2x² + c . . . . . (1)

Given that,

y\left( {{\pi \over 2}} \right) = 0

\therefore\,\,\,

x =

y\left( {{\pi \over 2}} \right) = 0

and y = 0

Put this x =

{\pi \over 2}

and y = 0 at equation (1)

0.1 = 2.

\left( {{\pi \over 2}} \right)

² + c

\Rightarrow \,\,\,

c =

- {{{\pi ^2}} \over 2}

So, differential equation is

y sin x = 2x²

-

{{{\pi ^2}} \over 2}\,\,.....(2)

Now we have to find y

\left( {{\pi \over 6}} \right).

So, put x =

{{\pi \over 6}}

at equation (2)

y . sin

{{\pi \over 6}}

= 2

{\left( {{\pi \over 6}} \right)^2} - {{{\pi ^2}} \over 2}

\Rightarrow \,\,\,\,\,\,\,y.{1 \over 2}

= 2.

{{{\pi ^2}} \over {36}} - {{{\pi ^2}} \over 2}

\Rightarrow \,\,\,\,\,{y \over 2} = {{{\pi ^2}} \over {18}} - {{{\pi ^2}} \over 2}

\Rightarrow \,\,\,\,{y \over 2} = {{{\pi ^2} - 9{\pi ^2}} \over {18}}

\Rightarrow \,\,\,\,\,y = - {{8{\pi ^2}} \over 9}