euler_hw - y ( x ) is, y ( x ) = A e i x + B e-i x (14) = C...

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Euler hw p.165 35, 36, 37(graded) September 29, 2010 I will do problem 34 and then use the results on the assigned hw. (34) Euler Equations - an equation of the form (2 nd or- der, can be extended to general form for any order), t 2 d 2 y dt 2 + α t dy dt + β y = 0 (1) (a)transform eq using substitution x = ln t to transform the equation to be a relation on x and y dy dt = dy dx dx dt (2) = 1 t dy dx (3) d 2 y dt 2 = d dt dy dt (4) = d dt ± dx dt dy dx ² (5) = d 2 x dt 2 dy dx + dx dt d dt ± dy dx ² (6) = d 2 x dt 2 dy dx + ± dx dt ² 2 d 2 y dx 2 (7) = - 1 t 2 dy dx + 1 t 2 d 2 y dx 2 (8) Substituting these definitions into the diff eq we find, ± d 2 y dx 2 - dy dx ² + α dy dx + β y = 0 (9) d 2 y dx 2 + ( α - 1) dy dx + β y = 0 (10) Since this equation is constant coeff we can assume a solution of the form y = e rx substituting it in we obtain the characteristic polynomial, r 2 + r ( α - 1) + β = 0 (11) which has roots, r = 1 - α 2 ± r α - 1 2 2 - β (12) (35) t 2 y 00 + t y 0 + y = 0 (13) From (12) with α = β = 1 we find r = ± i . The solution
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Unformatted text preview: y ( x ) is, y ( x ) = A e i x + B e-i x (14) = C cos x + D sin x (15) The nal solutions y ( t ) is, y ( t ) = C cos ln t + D sin ln t (16) (36) t 2 y 00 + 4 t y + 2 y = 0 (17) From (12) with = 4, = 2 we nd r =-2 ,-1. The solution y ( x ) is, y ( x ) = A e-2 x + B e-x (18) The nal solutions y ( t ) is, y ( t ) = A t 2 + B t (19) (37) t 2 y 00 + 3 t y + 1 . 25 y = 0 (20) From (12) with = 3, = 1 . 25 we nd r =-1 i/ 2. The solution y ( x ) is, y ( x ) = e-x A e i x/ 2 + B e-i x/ 2 (21) = e-x ( C cos x/ 2 + D sin x/ 2) (22) The nal solutions y ( t ) is, y ( t ) = 1 t C cos 1 2 ln t + D sin 1 2 ln t (23) 1...
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