Solutions To Linear Algebra Practice Problems Page 15
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18We can use the eigenvalues and eigenvectors computed in part (a).
In the basis of eigenvectors, the differential equation becomes:
u
2
0
u
1
1
=
u
0
1
u
2
2
So, we have the differential equations u
=
2u
and u
=
u
.
1
1
1
2
This has solutions
u
= A cos( 2t) + B sin( 2t)
1
u
= C cos t + D sin t
2
Thus:
y
1
3
1
= u
+ u
1
2
y
1
4
2
1
3
=
A cos( 2t) + B sin( 2t)
+ (C cos t + D sin t)
1
4
y
= A cos( 2t) + B sin( 2t) + 3C cos t + 3D sin t
1
y
= A cos( 2t) + B sin( 2t) + 4C cos t + 3D sin t
2
Note: Your answer may look different if you used a multiple of
the eigenvectors.
15
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