JEE Main 16 March 2021 Shift 1 Solved Paper

Given, ‌

dy

dx

+2y‌tan‌x=sin‌x
This differential equation is of the form ‌

dy

dx

+Py=Q where P and Q is function of x.
which is a linear differential equation.
Here, P=2‌tan‌x and Q=sin‌x
The integrating factor of linear differential equation is e∫Pdx.
Here, e∫2‌tan‌x‌d‌x=e∫‌

2‌sin‌x

cos‌x

dx=e−2‌log(cos‌x)=sec2x
Now, ‌

dy

dx

+2y‌tan‌x=sin‌x
On multiplying sec2× both the sides,
sec2x‌

dy

dx

+2ysec2x‌tan‌x=sin‌x‌sec2‌x
⇒‌‌‌

d

dx

(ysec2x)=sin‌x‌sec2‌x
⇒‌‌ysec2x=∫sin‌x‌sec2‌x‌d‌x
⇒‌‌ysec2x=∫‌

sin‌x

cos2x

dx
Let cos‌x=t
(−sin‌x)dx=dt
∫‌

−dt

t2

=‌

1

t

+c
So,ysec2x=‌

1

cos‌x

+c
ysec2x=sec‌x+c
Now, x=π∕3,y=0
0=2+c
⇒‌‌c=−2
So, ysec2x=sec‌x−2
y=cos2x(‌

1

cos‌x

−2)=−2cos2x+cos‌x
⇒‌‌y=−2(cos2x−‌

cos‌x

2

)
⇒‌‌y=−2[(cos2x−‌

cos‌x

2

+‌

1

16

)−‌

1

16

]
⇒‌‌y=−2[(cos‌x−1∕4)2−‌

1

16

]
⇒‌‌y=‌

1

8

−2(cos‌x−1∕4)2
So, ymin=‌

1

8