KCET 2015 Math Solved Paper

Given differential equations,
x‌

dy

dx

+2y=x2
⇒‌

dy

dx

+‌

2y

x

=x → (1)
General differential equation of this type is given by

dy

dx

+P(x)y=Q(x)
So, P=

2

x

then I.F. factor is given by
e∫Pdx=e∫

2

x

‌dx=e2‌log‌x=elog‌x2=x2
Then, x2‌

dy

dx

+2xy=x3
⇒‌

d

dx

(x2y)=x3
Integrating both the sides, we get
x2y=‌

x4

4

+C1
⇒x2y=‌

x4+C

4

⇒y=‌

x4+C

4x2