kreyszig 공업수학 解法(해법)9판 4장
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작성일 20-10-03 09:17
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y c1[ ] c2[ ]e0.024t.
on the right in the first equation and writing the system as a vector equation, we have
For t 0 this becomes, using the initial conditions y1(0) 0, y2(0) 150,
is five times that of the old T2. Ordering the system by interchanging the two terms
the new tank, because the limits are 25 lb and 125 lb, as the particular solution shows.
kreyszig 공업수학 解法(해법)9판 4장
y 25 [ ] 25 [ ]e0.024t.
this is 0.02x1 0.004x2 0.024x1. This simplifies to
설명
For 1 0 this is 0.02x1 0.004x2, say, x1 1, x2 5. For 20.024
y(0) [ ][ ]. Solution: c1 25, c2 25.
0.02x1 0.004x2 x1.
Hence a general solution of the system of ODEs is
2. The two balance equations (Inflow minus Outflow) change to
The situation described in the answer to Example 1 can no longer be achieved with
순서
This gives the particular solution
kreyszig 공업수학 솔루션9판 4장 kreyszig 공업수학 솔루션9판 4장 kreyszig 공업수학 솔루션9판 4장
kreyszig 공업수학 解法(해법)9판 4장
y2 0.02y1 0.004y2
of the vector equation Ax x; that is,
kreyszig 공업수학 솔루션9판 4장,ch4
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kreyszig 공업수학 解法(해법)9판 4장
The characteristic polynomial is 2 0.024 ( 0.024). Hence the eigenvalues
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kreyszig 공업수학 解法(해법)9판 4장
are 0 (as before) and 0.024. Eigenvectors can be obtained from the first component
0.004x1 0.004x2 0. A solution is x1 1, x2 1.
y1 0.004y2 0.02y1
where 0.004 appears because we divide through the content of the new tank, which
y Ay, where A [ ] .
다.
