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 uu U p u x U y u U q u x U y ± 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 ± c 1 y 1 U 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 u 2 u x U 0 U 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 u 2 is not defined). Hence y ± c 1 x 2 U c 2 x u 2 is a general solution of the given ODE. Set x ± 1 and use the initial conditions in y and y u ± 2 c 1 x u 2 c 2 x u 3 . This gives y u 1 U ± c 1 U c 2 ± 11, y u u 1 U ± 2 u c 1 u c 2 U ± u 6. The solution is obtained by inspection or elimination, c 1 ± 4, c 2 ± 7. 9 . Linear independence . e ax / e u ax ± e 2 ax is not constant, unless a ± 0. Hence y ± c 1 e ax U c 2 e u ax with a U 0 can be a general solution of an ODE. You may verify that the ODE is y uu u a 2 y ± 0. A derivation will be given in the next section. 11 . Linear dependence . This follows by noting that ln x 2 ± 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 u x , y , y u , y uu U ± 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 uu U y u 3 sin y ± 0 is Case (A). To reduce it, set z ± y u ± dy / dx and use y as the independent variable. This can be done by using the chain rule, namely, y uu ± dy u dx ± dy u dy ± dy dx ± dz dy z ± u z 3 sin y where the last equality follows by using the given ODE. Now divide by u z 3 and separate variables to get u dz z 2 ± sin y dy . Integration gives
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SSM-Ch02 - Chap. 2 Sec. 2.1 Second-Order Linear ODEs...

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