TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

Given differential equation,
sin‌x‌

dy

dx

+y‌cos‌x=e2x,x∈(0,π)
or ‌‌‌

dy

dx

+y‌cot‌x=‌

e2x

sin‌x

This is linear differential equation and general form is given by
‌

dy

dx

+P(x)y=Q(x)
Here P(x)=cot‌x,Q(x)=‌

e2x

sin‌x

Integrating factor, IF =e∫P(x)dx=e∫cot‌x‌d‌x=eln‌sin‌x
IF=sin‌x
Solution of linear differential equation
y(IF)=∫Q(x)IFdx
y(sin‌x)=∫‌

e2x

sin‌x

⋅sin‌x‌d‌x
y‌sin‌x=∫e2xdx⇒y‌sin‌x=‌

e2x

2

+C‌‌...(i)
\{where, C= integrating constant }
Given that, y(‌

π

2

)=0
Putting the value of c in Eq. (i), we get
y‌sin‌x=‌

e2x

2

−‌

eπ

2