Math322_HW1_Solution_ReviewODE

Math322_HW1_Solution_ReviewODE - Fall 2009: Math 322...

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Fall 2009: Math 322 Homework #1 Solution – ODE Review Lecturer: Boguk KIM ( kim@math.wisc.edu ) Room 415, Van Vleck Hall, Department of Mathematics, UW-Madison Distributed: Friday, October 2, 2009 1. (Review of ODE) This problem covers some solution methods for ODE. Although the listed solution methods are never comprehensive, please keep in mind that most solution methods to PDE are based on the ODE methods. Above all, note that there may be multiple solution methods for a diFerential equation. You do not have to present all possible solution methods in your homework, but it is advised that you consider other solution methods when you cannot get a desired solution with a speci±c method. You may want to verify your solutions with diFerent methods. And then, think about what would be the most eFective method in order to reach the desired solutions. (a) xy ± +2 y = x 3 ; y | x =1 =0 Classi±cation of ODE: linear ±rst-order ODE with variable coe²cients. Suggested method: method of integrating factor. Prototype of linear ±rst-order ODE: d y d x ( x )+ p ( x ) y ( x )= g ( x ) , (1.1) where x is the independent variable, y ( x ) is the unknown dependent variable, p ( x ) and g ( x ) are known and continuous functions on the connected domain where the initial condition is speci±ed. Applying the method of integrating factor: (i) Identify the prototype equation: p ( x 2 x ,g ( x x 2 . (1.2) (ii) Compute the integrating factor: μ ( x ) = e ± s p ( s )d s =e ± x 2 s d s = x 2 . (1.3) (iii) Multiply the integrating factor and use the product rule of diFerentiation to combine the ±rst two terms: ( μ ( x ) y ( x )) ± = μ ( x ) g ( x ) . (1.4) (iv) Integrate with respect to the independent variable: x 2 y ( x ² x x 0 s 2 · s 2 d s + y 0 = 1 5 ³ x 5 - 1 5 ´ +0 . (1.5) © 2009 Boguk Kim. Printed in Madison, WI. Math322_HW1_ReviewODE.tex; 12/10/2009; 6:10; p.1
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2 (v) Recover the solution y ( x ) by dividing the integrating factor: y ( x )= 1 5 ± x 3 - x - 2 ² . (1.6) (b) y ± - 2 y ± - 15 y = 0; y | x =0 =0 ,y ± | x =0 =1 ClassiFcation of ODE: linear homogeneous second-order ODE with constant coe±cients. Suggested method: method of characteristic equation. Prototype of linear homogeneous second-order ODE: a d 2 y d x 2 ( x )+ b d y d x ( x cy ( x ) = 0 . (1.7) where x is the independent variable, y ( x ) is the unknown dependent variable, a , b , c are constants. Applying the method of characteristic equation: (i) Try y ( x ) = e rx to Fnd the associated characteristic equation: ar 2 + br + c = 0 = r 2 - 2 r - 15 = 0 . (1.8) (ii) Solve the characteristic equation: ( r - 5)( r + 3) = 0 ⇐⇒ r =5 , - 3 . (1.9) (iii) Construct the general solution to the homogeneous equation: y ( x C 1 e 5 x + C 2 e - 3 x , (1.10) where C 1 and C 2 are arbitrary constants. (iv) Apply the initial conditions to determine C 1 and C 2 : y (0) = C 1 + C 2 , (1.11a) y ± ( x )=5 C 1 e 5 x - 3 C 2 e - 3 x = y ± (0) = 5 C 1 - 3 C 2 , (1.11b) (v) Compute the solution y ( x ): C 1 = 1 8 ,C 2 = - 1 8 , (1.12a) y ( x 1 8 ± e 5 x - e - 3 x ² . (1.12b) (c) xy ± - y ± = 1; y | x =1 ± | x =1 ClassiFcation of ODE: linear inhomogeneous second-order ODE with variable coe±cients.
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Math322_HW1_Solution_ReviewODE - Fall 2009: Math 322...

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