CGPET 2019 Solved Paper

ax2+by2=1
Differentiate w.r. to ' x', we get
a(2x)+b(2y)

dy

dx

=0 ⇒2[ax+by

dy

dx

]=0
⇒ax+by

dy

dx

=0
Again differentiate w.r. to ‌′x′, we get
[a+b(

dy

dx

)2+ by

d2y

dx2

]=0
⇒ a=−b(

dy

dx

)2−by

d2y

dx2

Put the value of a in the eq. (i), we get
[−b(

dy

dx

)2−by

d2y

dx2

]x +by

dy

dx

=0
⇒−b[{(

dy

dx

)2+y

d2y

dx2

}x−y

dy

dx

]=0
⇒[(

dy

dx

)2+y

d2y

dx2

]x−y

dy

dx

=0
This is the required differential equation and it is second order and first degree differential equation.