TS EAMCET 5 May 2018 Shift 2 Solved Paper

Consider the differential equation.
(

1

x2

+x)

dy

dx

+3y=1
It is solved as,
(

1+x3

x2

)

dy

dx

+3y=1

dy

dx

+

3x2

1+x3

y=

x2

1+x3

Calculate the integrating factor,
IF=e∫

3x2

1+x3

dx
=elog(1+x′)
=1+x3
Therefore, the solution to the differential equation is
y(1+x3)=∫(1+x3) (

x2

1+x3

)dx+k
y(1+x3)=

x3

3

+k
(3y−1)x3+3y=3k
(3y−1)x3+3y=c