Chapter2diffeq

# Chapter2diffeq - CHAPTER 2 Linearity and Nonlinearity 2.1...

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92 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 ± Linear Operation Notation 11. Using the common differential operator notation that D ( y ) = dy dt , we have the following: (a) 30 yt y y ′′ +−= can be written as L ( y ) = 0 for L = D 2 + tD 3. (b) y + y 2 = 0 is not a linear DE. (c) y + sin y = 1 is not a linear DE. (d) y + t 2 y = 0 can be written as L ( y ) = 0 for L = D + t 2 .

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SECTION 2.1 Linear Equations: The Nature of Their Solutions 93 (e) y + (sin t ) y = 1 can be written as L ( y ) = 1 for L = D + sin t . (f) 3 yy y ′′ −+ = sin t can be written as L ( y ) = sin t for L = D 2 3 D + 1. ± Linear and Nonlinear Operations 12. ( ) 2 Ly y y =+ Suppose 12 , and y are functions of t and c is any constant. Then () ( ) ()() () ( ) 11 2 2 2 22 2 y y Lc y c y c y c y c y cy Ly +=+ + + ++ = + = Hence, L is a linear operator. 13. ( ) 2 Ly y y To show that ( ) 2 is not linear we can pick a likely function of t and show that it does not satisfy one of the properties of linearity, equations (2) or (3). Consider the function yt = and the constant 5 c = : () () 2 2 55 2 5 5 5 Lt t t t tt t L t = + ⎛⎞ ≠+ = + = ⎜⎟ ⎝⎠ Hence, L is not a linear operator. 14. ( ) 2 t y Suppose and y are functions and c is any constant. ( ) ()() () ( ) 2 ty y y y y c y tc y cy t y cL y + + = + = 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.
94 CHAPTER 2 Linearity and Nonlinearity 15. ( ) t Ly y ey =− Suppose 12 , yy and y are functions of t and c is any constant. () ( ) () () ()() () 11 2 2 t tt Ly y y y e y y ye y ye y Ly Lc y c y e c y cy ey cL y +=+ + ′′ +− =+ = = 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 . 16. ( ) ( ) sin ty ( ) {} ()() ( ) 11 2 2 sin sin sin sin sin t y y yt y y y c y t c y cy cL y + + ++ = 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 . 17. ( ) ( ) 2 1 y y y =+− + { } 2 2 1 1 y c y c y y c y cL y + ≠+ + = Hence, ( ) is not a linear operator. ± Pop Quiz 18. 2 1 21 2 t y t c e += = + 19. ( ) 22 t y tc e ′ +=⇒ = + 20. ( ) 0.08 0.08 100 1250 t y t c e −= 21. 3 5 35 3 t y t c e ′ − =⇒ =

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SECTION 2.1 Linear Equations: The Nature of Their Solutions 95 22.
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## This note was uploaded on 04/07/2008 for the course MA 232 taught by Professor Toland during the Fall '08 term at Clarkson University .

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Chapter2diffeq - CHAPTER 2 Linearity and Nonlinearity 2.1...

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