t
y" (t) + -y' (t)
1 + t²
1 + t²
verify that y₁ (t) = √√1+ t² is a solution. Then use
Theorem 8 on page 198 to find y₂2 (t). Note that you may"
d
t
1
need to use the fact that
(1 + t²)³/2
·y(t) =
= 0. Directly
dt √1+t²
Then, find a particular solution to
t
1
y" (t) + ₁ + t²Y (t)
y'
1
variation of parameters.
1 + t2 y(t)
=
2
1 + t²
by the
Fig: 1