Find the general solution to the Cauchy-Euler equation
t
2
y
′
′
−
10
t
y
′
+
28
y
=
0
\displaystyle {t}^{{2}}{y}{''}-{10}{t}{y}'+{28}{y}={0}
t
2
y
′′
−
10
t
y
′
+
28
y
=
0
. Use
c
1
\displaystyle {c}_{{1}}
c
1
and
c
2
\displaystyle {c}_{{2}}
c
2
as arbitrary constants.
y
(
t
)
=
\displaystyle {y}{\left({t}\right)}=
y
(
t
)
=
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Question 6 Part 1 of 2
Next, find the solution that satisfies the initial conditions
y
(
1
)
=
−
1
,
y
′
(
1
)
=
−
2
\displaystyle {y}{\left({1}\right)}=-{1},{y}'{\left({1}\right)}=-{2}
y
(
1
)
=
−
1
,
y
′
(
1
)
=
−
2
.
y
(
t
)
=
\displaystyle {y}{\left({t}\right)}=
y
(
t
)
=
Preview
Question 6 Part 2 of 2
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\displaystyle
Enter your answer as an expression. Example: 3x^2+1, x/5, (a+b)/c
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Enter your answer as an expression. Example: 3x^2+1, x/5, (a+b)/c
Be sure your variables match those in the question