Tema25 - CHAPTER 25 25.1 The analytical solution can be...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHAPTER 25 25.1 The analytical solution can be derived by separation of variables dy y x dx ∫ ∫ =- 2 12 . ln . y x x C =- + 3 3 12 Substituting the initial conditions yields C = 0. Taking the exponential give the final result y e x x =- 3 3 1.2 The result can be plotted as 1 2 1 2 25.2 Euler’s method with h = 0.5 x y dy / dx 1-1.2 0.5 0.4-0.38 1 0.21-0.042 1.5 0.189 0.19845 2 0.288225 0.80703 Euler’s method with h = 0.25 gives x y dy / dx 1-1.2 0.25 0.7-0.79625 0.5 0.500938-0.47589 0.75 0.381965-0.2435 1 0.321089-0.06422 1.25 0.305035 0.110575 1.5 0.332679 0.349312 1.75 0.420007 0.782262 2 0.615572 1.723602 The results can be plotted along with the analytical solution as 1 2 1 2 25.3 For Heun’s method, the value of the slope at x = 0 can be computed as - 0.6 which can be used to compute the value of y at the end of the interval as y ( . ) ( . ( )) . . 0 5 1 12 1 0 5 0 4 = +- = The slope at the end of the interval can be computed as y '( . ) . ( . ) . ( . ) . 0 5 0 4 0 5 12 0 4 0 38 2 =- = - which can be averaged with the initial slope to predict y ( . ) . . . . 0 5 1 0 6 0 38 2 0 5 0 605 = +-- = This formula can then be iterated to yield j y i j ε a 0.4 1 0.605 33.9 2 0.5563124 8.75 3 0.5678757 2.036 4 0.5651295 0.4859 The remaining steps can be implemented with the result x i y i 0.5 0.5651295 1.0 0.4104059 1.5 0.5279021 2.0 2.181574 The results along with the analytical solution are displayed below: 1 2 1 2 25.4 The midpoint method with h = 0.5 x y dy / dx ym dy / dx-mid 1-1.2 0.7-0.79625 0.5 0.601875-0.57178 0.45893-0.29257 1 0.455591-0.09112 0.432812 0.156894 1.5 0.534038 0.56074 0.674223 1.255741 2 1.161909 3.253344 1.975245 7.629383 with h = 0.25 gives x y dy / dx ym dy / dx-mid 1-1.2 0.85-1.00672 0.25 0.74832-0.85121 0.641919-0.68003 0.5 0.578312-0.5494 0.509638-0.41249 0.75 0.47519-0.30293 0.437323-0.18996 1 0.4277-0.08554 0.417007 0.027366 1.25 0.434541 0.157521 0.454231 0.313703 1.5 0.512967 0.538615 0.580294 0.835986 1.75 0.721963 1.344657 0.890046 2.061012 2 1.237216 3.464206 1.670242 5.537897 The results can be plotted along with the analytical solution as 1 2 1 2 25.5 The 4 th-order RK method with h = 0.5 gives x y k1 ym k2 ym k3 ye k4 phi 1-1.2 0.7-0.79625 0.800938-0.91107 0.544467-0.51724-0.85531 0.5 0.572344-0.54373 0.436412-0.27821 0.50279-0.32053 0.412079-0.08242-0.30394 1 0.420375-0.08407 0.399356 0.144767 0.456567 0.165505 0.503128 0.528284 0.177459 1.5 0.509104 0.534559 0.642744 1.197111 0.808382 1.505611 1.26191 3.533348 1.578892 2 1.29855 3.635941 2.207535 8.526606 3.430202 13.24915 7.923127 40.01179 14.53321 1 2 1 2 25.6 ( a ) The analytical solution can be derived by separation of variables dy y x dx ∫ ∫ = + 1 2 2 2 y x x C = + + Substituting the initial conditions yields C = 2. Substituting this value and solving for y gives the final result y x x = + + ( ) 2 2 2 4 16 The result can be plotted as 2 4 0.5 1 ( b ) Euler’s method with h = 0.5 x y dy / dx 1 1 0.5 1.5 1.837117 1 2.418559 3.110343 Euler’s method with...
View Full Document

This note was uploaded on 03/13/2012 for the course AEROSPACE 301 taught by Professor Pfchang during the Spring '12 term at Shandong University.

Page1 / 20

Tema25 - CHAPTER 25 25.1 The analytical solution can be...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online