UPSC NDA Math Model Paper 4

x

dy

dx

+y=x3

dy

dx

+

y

x

=x2
It is form of

dy

dx

+Py=Q
Here, P=1∕x and Q=x2
I. F.=e∫pdx
I. F.=eln‌x=x
The solution of the linear equation is given by
y(x)=∫x2(x)dx+c
xy=

x

44

+c
Put initial condition x = 1 and y = 1 / 4 in the above equation we get ⇒ c = 0
xy=

x

4

y=

x

4