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Chapter 2

# Chapter 2 - CHAPTER 2 Linearity and Nonlinearity 2.1 Linear...

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CHAPTER 2 Linearity and Nonlinearity 2.1 Linear Equations: The Nature of Their Solutions ! Classification 1. First-order, nonlinear 2. First-order, linear, nonhomogeneous, variable coefficients 3. Second-order, linear, homogeneous, variable coefficients 4. Second-order, linear, nonhomogeneous, variable coefficients 5. Third-order, linear, homogeneous, constant coefficients 6. Third-order, linear, nonhomogeneous, constant coefficients 7. Second-order, linear, nonhomogeneous, variable coefficients 8. Second-order, nonlinear 9. Second-order, linear, homogeneous, variable coefficients 10. Second-order, nonlinear ! Pop Quiz 11. ′ + = ( ) = + y y y t ce t 2 2 12. ′ + = ( ) = + y y y t ce t 2 1 1 2 2 13. ′ − = ( ) = y y y t ce t 3 5 5 3 3 14. ′ − = ( ) = y y y t ce t 008 100 1250 0 08 . . 15. ′ + = y y 2 4 , y y t ce t 0 1 2 2 ( ) = ( ) = + , y c 0 1 1 ( ) = = − . Hence, y t e t ( ) = 2 2 . 16. ′ + = y y 5 1 , y y t ce t 1 0 1 5 5 ( ) = ( ) = + , y c e 1 0 1 5 5 ( ) = = − . Hence, y t e t ( ) = ( ) 1 5 1 5 1 b g . 69

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70 CHAPTER 2 Linearity and Nonlinearity ! Nonhomogeneous Principle I 17. The homogeneous equation is 2 0 x y = , which has solutions X h = ( ) α α α , arbitrary 2 k p and describes the line y x = 2 in the xy -plane. A particular solution of the nonhomogeneous equa- tion 2 5 x y = is X p = ( ) 0 5 , . Hence the solu- tion of 2 5 x y = consists of X = ( ) + ( ) { } 0 5 1 2 , , α 2 4 –2 –5 homogeneous solution general solution X p X h y x = 2 y x = 2 5 X X X = + h p where α is any real number; this solution is the parametric form of the line y x = 2 5 in the xy - plane. ! Superposition Principle 18. If y 1 and y 2 are solutions of ′ + ( ) = y p t y 0, then ′ + ( ) = ′ + ( ) = y p t y y p t y 1 1 2 2 0 0. Adding these equations gives ′ + ′ + ( ) + ( ) = y y p t y p t y 1 2 1 2 0 or y y p t y y 1 2 1 2 0 + + ( ) + = a f a f , which shows that y y 1 2 + is also a solution of the given equation. If y 1 is a solution, we have ′ + ( ) = y p t y 1 1 0 and multiplying by c we get c y p t y cy cp t y cy p t cy ′ + ( ) = ′ + ( ) = + ( ) = 1 1 1 1 1 1 0 0 0 a f a f a f , which shows that cy 1 is also a solution of the equation.
SECTION 2.1 Linear Equations: The Nature of Their Solutions 71 ! Second-Order Superposition Principle 19. If y 1 and y 2 are solutions of ′′ + ( ) ′ + ( ) = y p t y q t y 0, we have ′′+ ( ) ′ + ( ) = ′′+ ( ) ′ + ( ) = y p t y q t y y p t y q t y 1 1 1 2 2 2 0 0. Multiplying these equations by c 1 and c 2 respectively, then adding and using properties of the derivative, we arrive at c y c y p t c y c y q t c y c y 1 1 2 2 1 1 2 2 1 1 2 2 0 + ′′ + ( ) + ′ + ( ) + = a f a f a f , which shows that c y c y 1 1 2 2 + is also a solution. ! Linear and Nonlinear Operations 20. L y y y ( ) = ′ + 2 L cy cy cy cy cy c y y cL y ( ) = ( )′ + ( ) = ′ + = ′ + ( ) = ( ) 2 2 2 Hence, L is a linear operator. 21. L y y y ( ) = ′ + 2 L cy cy cy cy c y c y y cL y ( ) = ( )′ + ( ) = ′ + ′ + = ( ) 2 2 2 2 b g Hence, L is not a linear operator. 22. L y y ty ( ) = ′ + 2 L y y y y t y y y ty y ty L y L y L cy cy t cy c y ty cL y 1 2 1 2 1 2 1 1 2 2 1 2 2 2 2 2 2 + = + + + = ′ + + ′ + = + ( ) = ( )′ + ( ) = ′ + ( ) = ( ) a f a f a f a f a f a f a f Hence, L is a linear operator. This problem illustrates the fact that the coefficients of a DE can be functions of t and the operator will still be linear.

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Chapter 2 - CHAPTER 2 Linearity and Nonlinearity 2.1 Linear...

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