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285lect24 - 1 Lecture 25 Example 1(Continue 2y 00 3y 0 y =...

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1 Lecture 25 Example 1 (Continue) 2 y 00 + 3 y 0 + y = t 2 + 3 sin t: I. complementary solution. 2 y 00 + 3 y 0 + y = 0 : Characteristic equation: 2 r 2 + 3 r + 1 = 0 Solution is: r 1 = ° 1 ; r 2 ° 1 2 : y c ( t ) = C 1 e ° t + C 2 e ° t 2 : II. We see that the function t 2 + 3 sin t is not in the table. However, the functions f 1 ( x ) = t 2 and f 2 ( x ) = 3 sin t are listed in the table, see rows (a) and (c). So, we consider the following two problems: L [ y 1 p ] = t 2 L [ y 2 p ] = 3 sin t : III. Continue from this point. Set up solutions: r = 0 is not a root of the char. eq. Hence, s = 0 . So, by (a), r = i is not a root of the char. eq. Hence, s = 0 . So, by (c) y 1 p ( t ) = B 0 t 2 + B 1 t + B 2 y 2 p ( t ) = A cos t + B sin t 1
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IV. Write equations on coe¢ cients: y 0 1 p ( t ) = 2 B 0 t + B 1 y 0 2 p ( t ) = ° A sin t + B cos t y 00 1 p ( t ) = 2 B 0 y 00 2 p ( t ) = ° A cos t ° B sin t 2 y 00 + 3 y 0 + y = t 2 ) 2 y 00 + 3 y 0 + y = 3 sin t ) 2 (2 B 0 ) + 3 (2 B 0 t + B 1 ) + ( B 0 t 2 + B 1 t + B 2 ) = t 2 ) 2 ( ° A cos t ° B sin t ) +3 ( ° A sin t + B cos t ) + A cos t + B sin t = 3 sin t ) (4 B 0 + 3 B 1 + B 2 ) + t (6 B 0 + B 1 ) + t 2 B 0 = t 2 ) (3 B ° A ) cos t ° (3 A + B ) sin t = 3 sin t ) B 0 = 1 4 B 0 + 3 B 1 + B 2 = 0 6 B 0 + B 1 = 0 3 B ° A = 0 3 A +
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