kreyszig 공업수학 解法(해법)9판 4장
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작성일 21-06-18 08:05본문
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y(0) [ ][ ]. Solution: c1 25, c2 25.
The situation described in the answer to Example 1 can no longer be achieved with
2. The two balance equations (Inflow minus Outflow) change to
the new tank, because the limits are 25 lb and 125 lb, as the particular solution shows.
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
0.02x1 0.004x2 x1.
y c1[ ] c2[ ]e0.024t.
y2 0.02y1 0.004y2
순서
kreyszig 공업수학 솔루션9판 4장,ch4
kreyszig 공업수학 솔루션9판 4장 kreyszig 공업수학 솔루션9판 4장 kreyszig 공업수학 솔루션9판 4장
y 25 [ ] 25 [ ]e0.024t.
For t 0 this becomes, using the initial conditions y1(0) 0, y2(0) 150,
y Ay, where A [ ] .
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is five times that of the old T2. Ordering the system by interchanging the two terms
this is 0.02x1 0.004x2 0.024x1. This simplifies to
kreyszig 공업수학 解法(해법)9판 4장
kreyszig 공업수학 解法(해법)9판 4장
of the vector equation Ax x; that is,
설명
kreyszig 공업수학 解法(해법)9판 4장
Hence a general solution of the system of ODEs is
y1 0.004y2 0.02y1
0.004x1 0.004x2 0. A solution is x1 1, x2 1.
레포트 > 공학,기술계열
The characteristic polynomial is 2 0.024 ( 0.024). Hence the eigenvalues
This gives the particular solution
on the right in the first equation and writing the system as a vector equation, we have
kreyszig 공업수학 解法(해법)9판 4장
For 1 0 this is 0.02x1 0.004x2, say, x1 1, x2 5. For 20.024
다.