Solve:
y
′
′
−
3
y
′
+
2
y
=
6
t
−
7
\displaystyle {y}{''}-{3}{y}'+{2}{y}={6}{t}-{7}
y
′′
−
3
y
′
+
2
y
=
6
t
−
7
y
(
0
)
=
1
,
y
′
(
0
)
=
1
\displaystyle {y}{\left({0}\right)}={1},\ {y}'{\left({0}\right)}={1}
y
(
0
)
=
1
,
y
′
(
0
)
=
1
y
(
t
)
\displaystyle {y}{\left({t}\right)}
y
(
t
)
=
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