kreyszig 공업수학 해법9판 4장
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작성일 21-10-17 20:16
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For t 0 this becomes, using the initial conditions y1(0) 0, y2(0) 150,
where 0.004 appears because we divide through the content of the new tank, which
설명
y 25 [ ] 25 [ ]e0.024t.
y c1[ ] c2[ ]e0.024t.
레포트 > 공학,기술계열
kreyszig 공업수학 솔루션9판 4장,ch4
For 1 0 this is 0.02x1 0.004x2, say, x1 1, x2 5. For 20.024
kreyszig 공업수학 해법9판 4장
The characteristic polynomial is 2 0.024 ( 0.024). Hence the eigenvalues
kreyszig 공업수학 해법9판 4장
y Ay, where A [ ] .
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The situation described in the answer to Example 1 can no longer be achieved with
are 0 (as before) and 0.024. Eigenvectors can be obtained from the first component
순서
this is 0.02x1 0.004x2 0.024x1. This simplifies to
is five times that of the old T2. Ordering the system by interchanging the two terms
Hence a general solution of the system of ODEs is
kreyszig 공업수학 해법9판 4장
y2 0.02y1 0.004y2
kreyszig 공업수학 솔루션9판 4장 kreyszig 공업수학 솔루션9판 4장 kreyszig 공업수학 솔루션9판 4장
This gives the particular solution
0.004x1 0.004x2 0. A solution is x1 1, x2 1.
the new tank, because the limits are 25 lb and 125 lb, as the particular solution shows.
y(0) [ ][ ]. Solution: c1 25, c2 25.
y1 0.004y2 0.02y1
of the vector equation Ax x; that is,
kreyszig 공업수학 해법9판 4장
on the right in the first equation and writing the system as a vector equation, we have
2. The two balance equations (Inflow minus Outflow) change to
0.02x1 0.004x2 x1.
다.
