{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–2. Sign up to view the full content.

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 ) = 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 ) . 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 dx 2 = f ( x , y , y 0 ), y ( x 0 ) = y 0 , y 0 ( x 0 ) = y 1 . Existence and Uniqueness Theorem (EUT) : Given the IVP 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 = φ( 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 dt + 3 θ = - 2 e 2 t . Solution Now d θ/ dt = 6 e 3 t - 2 e 2 t and d 2 θ/ dt 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 dy dx = 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 dy / dx . y - ln y = x 2 + 1 y 0 - 1 y y 0 = 2 x y 0 = 2 x ( 1 - 1 y ) dy dx = 2 xy y - 1 Accordingly, the relationship is an implicit solution of the DE. 14/14 Show that φ( 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 φ( x ) = c 1 sin x + c 2 cos x is an explicit solution of the DE for any constants c 1 , c 2 . We say it constitutes a

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}