Appendix

# Appendix - APPENDIX A EXISTENCE AND UNIQUENESS OF SOLUTIONS...

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588 Appendix A APPENDIX A EXISTENCE AND UNIQUENESS OF SOLUTIONS In Problems 1–12 we apply the iterative formula 1 (, () ) x nn a yb f t y t d t + =+ to compute successive approximations { y n ( x )} to the solution of the initial value problem y = f ( x , y ), y ( a ) = b . starting with y 0 ( x ) = b . 1 . y 0 ( x ) = 3 y 1 ( x ) = 3 + 3 x y 2 ( x ) = 3 + 3 x + 3 x 2 /2 y 3 ( x ) = 3 + 3 x + 3 x 2 /2 + x 3 y 4 ( x ) = 3 + 3 x + 3 x 2 x 3 x 4 /8 y ( x ) = 3 - 3 x + 3 x 2 x 3 x 4 /8 + ⋅⋅⋅ = 3 e x 2. y 0 ( x ) = 4 y 1 ( x ) = 4 - 8 x y 2 ( x ) = 4 - 8 x + 8 x 2 y 3 ( x ) = 4 - 8 x + 8 x 2 - (16/3) x 3 y 4 ( x ) = 4 - 8 x + 8 x 2 - (16/3) x 3 + (8/3) x 4 y ( x ) = 4 - 8 x + 8 x 2 - (16/3) x 3 + (8/3) x 4 - ⋅⋅⋅ = 4 e - 2 x 3. y 0 ( x ) = 1 y 1 ( x ) = 1 - x 2 y 2 ( x ) = 1 - x 2 + x 4 y 3 ( x ) = 1 - x 2 + x 4 - x 6 /6 y 4 ( x ) = 1 - x 2 + x 4 - x 6 /6 + x 8 /24

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Appendix A 589 y ( x ) = 1 - x 2 + x 4 /2 - x 6 /6 + x 8 /24 - ⋅⋅⋅ = exp( - x 2 ) 4. y 0 ( x ) = 2 y 1 ( x ) = 2 + 2 x 3 y 2 ( x ) = 2 + 2 x 3 + x 6 y 3 ( x ) = 2 + 2 x 3 + x 6 + (1/3) x 9 y 4 ( x ) = 2 + 2 x 3 + x 6 + (1/3) x 9 + (1/12) x 12 y ( x ) = 2 + 2 x 3 + x 6 + (1/3) x 9 + (1/12) x 12 + ⋅⋅⋅ = 2 exp( x 3 ) 5. y 0 ( x ) = 0 y 1 ( x ) = 2 x y 2 ( x ) = 2 x + 2 x 2 y 3 ( x ) = 2 x + 2 x 2 + 4 x 3 /3 y 4 ( x ) = 2 x + 2 x 2 + 4 x 3 /3 + 2 x 4 y ( x ) = 2 x + 2 x 2 + 4 x 3 x 4 /3 + ⋅⋅⋅ = e 2 x - 1 6. y 0 ( x ) = 0 y 1 ( x ) = (1/2) x 2 y
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Appendix - APPENDIX A EXISTENCE AND UNIQUENESS OF SOLUTIONS...

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