L12 - Fall 2003 Math 308/501502 1 Introduction 1.2...

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Fall 2003 Math 308/501–502 1 Introduction 1.2 Solutions & Initial Value Problems Wed, 03/Sep c ± 2003, Art Belmonte Summary A general form for an n th-order ODE is F ± x , y , y 0 , y 00 ,... y ( n ) ² = 0 . Here x is the independent variable and y is the dependent variable. This equation is assumed to hold on some open real interval I , possibly of infinite extent. Preferably, we can isolate y ( n ) ,the n th derivative of y ,and rewrite the ODE in the form y ( n ) = f ± x , y , y 0 , y 00 ,..., y ( n - 1 ) ² . An explicit solution is a function y = φ( x ) which when substituted into of one of the aforementioned ODEs yields an identity for all x in I . This is shown via differentiation and direct substitution. An implicit solution is a relation G ( x , y ) = 0 which yields one or more explicit solutions on I , perhaps only theoretically (say via the Implicit Function Theorem from mulitivariable calculus or advanced calculus). This is usually shown via implicit differentiation. An n th-order initial value problem (IVP) consists of an n th-order differential equation together with n initial conditions (ICs) y ( k ) ( x 0 ) = y k , k = 0 , 1 n - 1 . Here x 0 and the y k are constants and y ( k ) denotes the k th derivative of y , with y ( 0 ) signifying y itself. Therefore, a first-order IVP is usually specified as y 0 = dy dx = f ( x , y ), y ( x 0 ) = y 0 . Similarly, a second-order IVP is given as y 00 = d 2 y 2 = f ( x , y , y 0 ), y ( x 0 ) = y 0 , y 0 ( x 0 ) = y 1 . Existence and Uniqueness Theorem (EUT) :GiventheIVP y 0 = f ( x , y ), y ( x 0 ) = y 0 ,let f and f y = f /∂ y be continuous on R = { ( x , y ) : a < x < b , c < y < d } , an open rectangle containing ( x 0 , y 0 ) . Then the IVP has a unique solution y = x ) on J = ( x 0 - δ< x < x 0 + δ) , an open interval centered at x 0 . Hand Examples 14/6 Determine whether the function θ = 2 e 3 t - e 2 t is a solution of the differential equation d 2 θ dt 2 - θ d θ + 3 θ =- 2 e 2 t . Solution Now d θ/ = 6 e 3 t - 2 e 2 t and d 2 2 = 18 e 3 t - 4 e 2 t . Substituting into the DE yields ( 18 e 3 t - 4 e 2 t ) - ( 2 e 3 t - e 2 t )( 6 e 3 t - 2 e 2 t ) + 3 ( 2 e 3 t - e 2 t ) 2 e 2 t - 12 e 6 t + 10 e 5 t - 2 e 4 t + 24 e 3 t - 9 e 2 t = 0 an equation which holds on no open interval I R . Therefore, θ is not a solution of the differential equation. NOTE: The fact that the last equation actually holds at one isolated point ( t 0 . 36) is insufficient. It must hold over an entire open interval. 14/10 Determine whether the relation y - ln y = x 2 + 1 is an implicit solution of the differential equation = 2 xy y - 1 . Solution Via the Implicit Function Theorem, we assume that the relationship does indeed define y as a function of x . Implicitly differentiate the relation with respect to x , then isolate / . y - ln y = x 2 + 1 y 0 - 1 y y 0 = 2 x y 0 = 2 x ( 1 - 1 y ) = 2 y - 1 Accordingly, the relationship is an implicit solution of the DE. 14/14 Show that x ) = c 1 sin x + c 2 cos x is an explicit solution to y 00 + y = 0 for any constants c 1 , c 2 . Solution Now y 0 = φ 0 ( x ) = c 1 cos x - c 2 sin x and y 00 = φ 00 ( x ) c 1 sin x - c 2 cos x . Substituting into the DE yields ( - c 1 sin x - c 2 cos x ) + ( c 1 sin x + c 2 cos x ) = 0 , an identity for all real x . Therefore x ) = c 1 sin x + c 2 cos x is an explicit solution of the DE for any constants c 1 , c 2 .Wesayi t constitutes a 2-parameter family of solutions to the DE. The
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This note was uploaded on 03/29/2008 for the course MATH 308 taught by Professor Comech during the Spring '08 term at Texas A&M.

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L12 - Fall 2003 Math 308/501502 1 Introduction 1.2...

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