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The last formula implies there exist constant vectors A 3 and A 4 such that (531) Q 2 (y 3) = A 3 cos ⁡ (c y 3) A 4 sin ⁡ (c y 3) Therefore the general solutions of system are (532) x = e c y 1 (A 2 cosh ⁡ (c y 2) A 3 cos ⁡ (c y 3) A 4 sin ⁡ (c y 3) sinh ⁡ (c y 2)) A 1, where A i are constant vectorsThe triangle JOY is translated to another triangle which each points is moved to the right by 3 and moved downward by 2 So if a two segments is connected of two of each points, the result is that they have the same length and they are parallel to each otherH L O l) l) O 24 H & D & z z 0 Z D & L O < L O D O z D l) < L D 0 3 2 0 D H G )V l) L L 0 Z D < O L & 0 D & & L O) < D 1 5 0 o D < L _fil & < z t) L & & & & l) It z < & ž 0 Q) D a o rEïJ D ò 7 2 4 Z o 1 6 D a ò D o åJËJ D 19 ò fiJ IC O D < & O 25 D D Title SKM_C

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