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Uae c cxg t. A N Z X } b v B F s { ́B C X g ^ { f B P A z y W B { i J C v N e B b N ƃ{ f B P A E t P A E G X e E p h i W ̂ X ł B @ F s { s h 318 @ 啟 r PF B waisu@quartzocnnejp. G X e T @ A l J v A B S \ @ v C x g T i P T l j B n C A G X e ̍ō ̖ Ȃ ɁB TwoRings Ёi E y j @ 1000 `20 j Y G X e ͂. Y(t) = C e 2t, for any constant C That is, any function of this form, regardless of the value of C, will satisfy the equation y′ − 2 y = 0 While there are infinitely many such functions, no other type of functions could satisfy the equation The similar technique could also be used to solve this next example.
O _ C i R A c SD K _ ^ T C g u v k @ r c f t m c ` l v. N u c Y @ c A C x g u z K ΐV ԉΑ ƐM B ӍՁI v y z @ @ @ @ 쌧 z K S x m x m Ԓn. R ` t @ N g I C X g A.
T PMS C x g uSPRING 9KEYS ROAD 11 v ͂ I Imp No W ^ C g A e B X g 1 1. 13 F00 `14 F00 u Q n h f E X N 2 v C X g N ^ @ c T t A ؎ V O n h o ̂Ȃ A O Ɍ ẴX N ł B. I B G E A T C X ē f w N x ̂c u c Ȃǂ̒ʔ̍ŐV J Ă ܂ B I B G E A T C X ē f w N x ̂c u c ʔ̂Ȃ A T C g ɂ C B.
I @ 茧 ދn S I g @TEL F @ @. C C′ This looks reasonable purely formally, since we get the right side by substituting into the left side the expressions for z and dz in terms of x and y z = f(x, y), dz = fxdx fydy To justify it more carefully, we use the parametrizations given above for C and C′ to calculate the line integrals t1 dz P (x, y, z)dz = P (x(t),y(t),z(t.
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