Consider the linear system :
d
x
d
t
=
−
2
x
−
2
y
\displaystyle \frac{{{\left.{d}{x}\right.}}}{{\left.{d}{t}\right.}}=-{2}{x}-{2}{y}
d
t
d
x
=
−
2
x
−
2
y
d
y
d
t
=
−
2
x
+
y
\displaystyle \frac{{{\left.{d}{y}\right.}}}{{\left.{d}{t}\right.}}=-{2}{x}+{y}
d
t
d
y
=
−
2
x
+
y
with initial conditions
x
(
0
)
=
0
and
y
(
0
)
=
1
\displaystyle {x}{\left({0}\right)}={0}{\quad\text{and}\quad}{y}{\left({0}\right)}={1}
x
(
0
)
=
0
and
y
(
0
)
=
1
. Solve this IVP and enter the formulas for the component functions below.
x
(
t
)
=
\displaystyle {x}{\left({t}\right)}=
x
(
t
)
=
Preview
Question 6 Part 1 of 2
y
(
t
)
=
\displaystyle {y}{\left({t}\right)}=
y
(
t
)
=
Preview
Question 6 Part 2 of 2
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\displaystyle