kreyszig 공업수학 해결책9판 4장
페이지 정보
작성일 21-10-08 17:30본문
Download : ch04.pdf
of the vector equation Ax x; that is,
The situation described in the answer to Example 1 can no longer be achieved with
on the right in the first equation and writing the system as a vector equation, we have
y c1[ ] c2[ ]e0.024t.
0.004x1 0.004x2 0. A solution is x1 1, x2 1.
This gives the particular solution
For 1 0 this is 0.02x1 0.004x2, say, x1 1, x2 5. For 20.024
is five times that of the old T2. Ordering the system by interchanging the two terms
kreyszig 공업수학 해결책9판 4장
y Ay, where A [ ] .
the new tank, because the limits are 25 lb and 125 lb, as the particular solution shows.
kreyszig 공업수학 해결책9판 4장
설명
kreyszig 공업수학 해결책9판 4장
순서
레포트 > 공학,기술계열
Download : ch04.pdf( 86 )
where 0.004 appears because we divide through the content of the new tank, which
are 0 (as before) and 0.024. Eigenvectors can be obtained from the first component
kreyszig 공업수학 솔루션9판 4장 kreyszig 공업수학 솔루션9판 4장 kreyszig 공업수학 솔루션9판 4장
y(0) [ ][ ]. Solution: c1 25, c2 25.
y1 0.004y2 0.02y1
kreyszig 공업수학 해결책9판 4장
this is 0.02x1 0.004x2 0.024x1. This simplifies to
Hence a general solution of the system of ODEs is
2. The two balance equations (Inflow minus Outflow) change to
kreyszig 공업수학 솔루션9판 4장,ch4
y2 0.02y1 0.004y2
For t 0 this becomes, using the initial conditions y1(0) 0, y2(0) 150,
The characteristic polynomial is 2 0.024 ( 0.024). Hence the eigenvalues
0.02x1 0.004x2 x1.
y 25 [ ] 25 [ ]e0.024t.
다.