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SSM-Ch02 - Chap 2 Sec 2.1 Second-Order Linear ODEs...

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Chap . 2 Second - Order Linear ODEs Sec . 2 . 1 Homogeneous Linear ODEs of Second Order On pp. 45-46 we extend concepts defined in Chap. 1, notably solution and homogeneous and nonhomogeneous, to second-order ODEs; take a look into Secs. 1.1 and 1.5 before you continue. You will see in this section that a homogeneous linear ODE is of the form y u2032u2032 u002b p ue0a2 x ue0a3 y u2032 u002b q ue0a2 x ue0a3 y u003d 0. An initial value problem for it will consist of two conditions, prescribing an initial value and an initial slope of the solution, both at the same point x 0 . But on the other hand, a general solution will now involve two arbitrary constants for which some values can be determined from the two initial conditions. Indeed, a general solution is of the form y u003d c 1 y 1 u002b c 2 y 2 where y 1 and y 2 are such that they cannot be pooled together with just one arbitrary constant remaining; expressed technically, y 1 and y 2 are “linearly independent”, meaning that they are not proportional on the interval on which a solution of the initial value problem is sought. Problem Set 2 . 1 . Page 52 5 . General solution . Initial value problem . Substitution shows that x 2 and x u2212 2 ue0a2 x u2260 0 ue0a3 are solutions. (The simple algebraic derivation of such solutions will be shown in Sec. 2.5 on p. 70.) They are linearly independent (not proportional) on any interval not containing 0 (where x u2212 2 is not defined). Hence y u003d c 1 x 2 u002b c 2 x u2212 2 is a general solution of the given ODE. Set x u003d 1 and use the initial conditions in y and y u2032 u003d 2 c 1 x u2212 2 c 2 x u2212 3 . This gives y ue0a2 1 ue0a3 u003d c 1 u002b c 2 u003d 11, y u2032 ue0a2 1 ue0a3 u003d 2 ue0a2 c 1 u2212 c 2 ue0a3 u003d u2212 6. The solution is obtained by inspection or elimination, c 1 u003d 4, c 2 u003d 7. 9 . Linear independence . e ax / e u2212 ax u003d e 2 ax is not constant, unless a u003d 0. Hence y u003d c 1 e ax u002b c 2 e u2212 ax with a u2260 0 can be a general solution of an ODE. You may verify that the ODE is y u2032u2032 u2212 a 2 y u003d 0. A derivation will be given in the next section. 11 . Linear dependence . This follows by noting that ln x 2 u003d 2 ln x . The problem is typical of cases in which some functional relation is used to show linear dependence. Problem 13 is of the same kind. 21 . Reduction to first order . The most general second-order ODE is of the form F ue0a2 x , y , y u2032 , y u2032u2032 ue0a3 u003d 0 . It can be reduced to first order if [Case (A)] x does not occur explicitly, or if [Case B] y does not occur explicitly. The ODE y u2032u2032 u002b y u2032 3 sin y u003d 0 is Case (A). To reduce it, set z u003d y u2032 u003d dy / dx and use y as the independent variable. This can be done by using the chain rule, namely, y u2032u2032 u003d dy u2032 dx u003d dy u2032 dy u2219 dy dx u003d dz dy z u003d u2212 z 3 sin y where the last equality follows by using the given ODE. Now divide by u2212 z 3 and separate variables to get u2212 dz z 2 u003d sin y dy .
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