Solve the initial value problem below.
x
2
y
′
′
−
3
x
y
′
+
5
y
=
0
,
y
(
1
)
=
2
,
y
′
(
1
)
=
−
1
\displaystyle {x}^{{2}}{y}{''}-{3}{x}{y}'+{5}{y}={0},\quad{y}{\left({1}\right)}={2},\quad{y}'{\left({1}\right)}=-{1}
x
2
y
′′
−
3
x
y
′
+
5
y
=
0
,
y
(
1
)
=
2
,
y
′
(
1
)
=
−
1
y
\displaystyle {y}
y
=
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