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Differential Equations of Second Order

# Differential Equations of Second Order - 916 CHAPTER 17...

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916 CHAPTER 17 Ordinary Differential Equations (a) Show that the solution of (B) remains between the solutions of (A) and (C) on any interval [0 , X ]where solutions of all three problems exist. Hint: We must have u ( x ) 1, y ( x ) 1, and v( x ) 1on[0 , X ]. (Why?) Apply the result of Exercise 16 to φ = y u and to φ = v y . (b) Find explicit solutions for problems (A) and (C). What can you conclude about the solution to problem (B)? (c) H I Use the Runge–Kutta method with h = 0 . 05, h = 0 . 02, and h = 0 . 01 to approximate the solution to (B) on [0 , 1]. What can you conclude now? 17.4 Differential Equations of Second Order The general second-order ordinary differential equation is of the form F ± d 2 y dx 2 , dy , y , x ² = 0 for some function F of four variables. When such an equation can be solved explicitly for y as a function of x , the solution typically involves two integrations and therefore two arbitrary constants. A unique solution usually results from prescribing the values of the solution y and its ±rst derivative y 0 = / at a particular point. Such a prescription constitutes an initial-value problem for the second-order equation. Equations Reducible to First Order A second-order equation of the form F ± d 2 y 2 , , x ² = 0 that does not involve the unknown function y explicitly (except through its derivatives) can be reduced to a ±rst-order equation by a change of dependent variable; if v = / , then the equation can be written F ± d v ,v, x ² = 0 . This ±rst-order equation in v may be amenable to the techniques described in earlier sections. If an explicit solution v = x ) can be found and integrated, then the function y = Z x ) is an explicit solution of the given equation. Example 1 Solve the initial-value problem d 2 y 2 = x ± ² 2 , y ( 0 ) = 1 , y 0 ( 0 ) =− 2 . Solution If we let v = / , the given differential equation becomes d v = x v 2 , which is a separable ±rst-order equation. Thus d v v 2 = xdx 1 v = x 2 2 + C 1 2 v 2 x 2 + C 1 .

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SECTION 17.4: Differential Equations of Second Order 917 The initial condition y 0 ( 0 ) =− 2 implies that v( 0 ) 2andso C 1 = 1 . Therefore, y 2 Z dx x 2 + 1 2tan 1 x + C 2 .
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Differential Equations of Second Order - 916 CHAPTER 17...

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