In section 3.5, read pages 315-319. -2 1 the repeated eigenvalue is A - 2 (A) For the example where A = -2 (B) If the initial condition for this system is Y (0) = (3, 0), then Vo and V1 Note: enter vectors horizontally surrounded by parentheses! Furthermore, the solution to the IVP would be Y(t)=( 3e 2t

Advanced Engineering Mathematics
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ISBN:9780470458365
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Chapter2: Second-order Linear Odes
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Could someone please explain this for me? 

In section 3.5, read pages 315-319.
-2
1
(A) For the example where A =
the repeated eigenvalue is A = -2
- 2
(B) If the initial condition for this system is Y (0) = (3, 0), then
Vo
and V1
Note: enter vectors horizontally surrounded by parentheses!
Furthermore, the solution to the IVP would be
Ý (t)=( 3e 2t
V, 0
C) If the initial condition for this system is Y (0) = (6, 3), then
Vo =
and V,
and the solution to the IVP would be
Y(t)=( 6e-2t + 3te 2t
V 3e 2t
Transcribed Image Text:In section 3.5, read pages 315-319. -2 1 (A) For the example where A = the repeated eigenvalue is A = -2 - 2 (B) If the initial condition for this system is Y (0) = (3, 0), then Vo and V1 Note: enter vectors horizontally surrounded by parentheses! Furthermore, the solution to the IVP would be Ý (t)=( 3e 2t V, 0 C) If the initial condition for this system is Y (0) = (6, 3), then Vo = and V, and the solution to the IVP would be Y(t)=( 6e-2t + 3te 2t V 3e 2t
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