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. +=⇒ () =+ yy y tc e t 22 12. += () y t c e t 21 1 2 2 13. −= =− y t c e t 35 5 3 3 14. y t c e t 008 100 1250 . . 15. 24 , t c e t 01 2 2 =⇒ () , yc 1 =⇒=− . Hence, yt e t 2 2 . 16. 51 , t c e t 10 1 5 5 , e 1 5 5 =⇒= . Hence, e t −− 1 5 1 bg . 69

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70 CHAPTER 2 Linearity and Nonlinearity ! Nonhomogeneous Principle I 17. The homogeneous equation is 20 xy −= , which has solutions X h = () ααα , arbitrary 2 kp and describes the line yx = 2 in the xy -plane. A particular solution of the nonhomogeneous equa- tion 25 is X p =− 05 . Hence the solu- tion of consists of X + {} 1 2 ,, α 24 –2 –5 homogeneous solution general solution X p X h = 2 XX X =+ hp where is any real number; this solution is the parametric form of the line in the xy - plane. ! Superposition Principle 18. If y 1 and y 2 are solutions of ′ + () = yp t y 0, then = = t y t y 11 22 0 0. Adding these equations gives ′ + + () = yyp t t y 12 1 2 0 or yy p t yy 0 + += afaf , which shows that + is also a solution of the given equation. If y 1 is a solution, we have = t y 0 and multiplying by c we get cy pty cy cp t y cy p t cy = ′ + = = 0 0 0 af 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 ′′ + () ′ + () = yp t y q t y 0, we have ′′+ () ′ + () = = t yq t y t t y 111 222 0 0. Multiplying these equations by c 1 and c 2 respectively, then adding and using properties of the derivative, we arrive at cy cy pt cy cy qt cy cy 11 22 0 + ′′ + () + ′ + () += a f a f a f , which shows that cy 2 2 + is also a solution. ! Linear and Nonlinear Operations 20. Ly y y () = ′ + 2 Lcy y cLy () = ′ + = ′ + = ′ + = () 2 Hence, L is a linear operator. 21. y y ′ + 2 c y y cL y = ′ + = ′ + ′ + 2 2 bg Hence, L is not a linear operator. 22. y ty ′ + 2 y y y y yt y y tcy cL y 12 1122 2 +=+ ++ = ′ + =+ = ′ + = ′ + a f a f a f af a f af af 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. 23. y ey t ′ − y y y e y y ye y ye y e cy ey cL y t tt −+ = ′ − = ′ − = ′ − b g

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72 CHAPTER 2 Linearity and Nonlinearity Hence, L is a linear operator. This problem illustrates the fact that a linear operator need not have coefficients that are linear functions of t .
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## This lab report was uploaded on 04/04/2008 for the course APPM 2360 taught by Professor Williamheuett during the Fall '07 term at Colorado.

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

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