Explanation Procedure used: Method of undetermined coefficients: A particular solution to the differential equation a y ″ + b y ′ + c y = C t m e r t , where m is a nonnegative integer, is of the form y p ( t ) = t s ( A m t m + ⋯ + A 1 t + A 0 ) e r t with: (i) s = 0 , if r is not a root of the associated auxiliary equation; (ii) s = 1 , if r is a simple root of the associated auxiliary equation; (iii) s = 2 , if r is a double root of the associated auxiliary equation. Calculation: The given differential equation is, 10 y ″ + y ′ − 3 y = t − e t 2 . The corresponding auxiliary equation is 10 r 2 + r − 3 = 0 . Simplify 10 r 2 + r − 3 = 0 as, 10 r 2 + 6 r − 5 r − 3 = 0 2 r ( 5 r + 3 ) − 1 ( 5 r + 3 ) = 0 ( 2 r − 1 ) ( 5 r + 3 ) = 0 r = 1 2 , − 3 5 That is, the roots are r 1 = 1 2 and r 2 = − 3 5 . Thus, the general solution for the homogenous equation is y ( t ) = c 1 e t 2 + c 2 e − 3 2 t . The RHS of the differential equation 10 y ″ + y ′ − 3 y = t − e t 2 has two terms t and − e t 2 . Comparing the term e t 2 with the form C t m e r t , it is observed that m = 0 and r = 1 2 . Since r = 1 2 is a root of the associated auxiliary equation, s = 1 . Thus, the particular solution form y p ( t ) = t s ( A m t m + ⋯ + A 1 t + A 0 ) e r t for s = 1 , m = 0 and r = 1 2 will be, y p ( t ) = t 1 ( A 0 ) e t 2 = A 0 t e t 2 Similarly, comparing the term t with the form C t m e r t , it is observed that m = 1 and r = 0 . Since r = 0 is not a root of the associated auxiliary equation, s = 0 . Thus, the particular solution form y p ( t ) = t s ( A m t m + ⋯ + A 1 t + A 0 ) e r t for s = 0 , m = 1 and r = 0 will be, y p ( t ) = t 0 ( A 1 t + A 0 ) e 0 t = A 1 t + A 0 Combine the particular solution forms for the individual terms and rename the coefficients as, y p ( t ) = A t e t 2 + B t + C . Obtain the first and second order derivatives of y p ( t ) = A t e t 2 + B t + C as follows: Differentiate y p ( t ) = A t e t 2 + B t + C with respect to t as, y ′ p ( t ) = 1 2 A t e t 2 + A e t 2 + B = ( t 2 + 1 ) A e t 2 + B

Fundamentals of Differential Equations (9th Edition)

9th Edition

ISBN: 9780321977069

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

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>tt 1:32 > trend.1m 1m (sales > summary(trend.1m) - tt) #3###23 (i) #### Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 2107.220 57.997 36.332e-16 *** tt -43.500 3.067 -14.18 7.72e-15 *** > trend = ts (fitted (trend.1m), start-start (sales), freq-frequency (sales)) sales trend ###23%23 (ii) #### as.numeric((1:32 %% 4) > X > q1 > q2 > q3 > 94 = = = = - as.numeric((1:32 %% 4) as.numeric((1:32 %% 4) as.numeric((1:32 %% 4) == 1) 2) == == 3) == 0) > season.lm = 1m (resid (trend.1m) 0+q1 + q2 + q3 + q4) #3##23%23 (iii) #### > summary(season.1m) Coefficients: Estimate Std. Error t value Pr(>|t|) q1 -38.41 43.27 -0.888 0.38232 92 18.80 43.27 0.435 0.66719 q3 -134.78 43.27 -3.115 0.00422 ** 94 154.38 43.27 3.568 0.00132 ** > season = ts (fitted (season.lm), start=start (sales), freq=frequency (sales)) > Y X season %23%23%23%23 (iv) #### >ar (Y, aic=FALSE, order.max=1) #23%23%23%23 (v) #### Coefficients: 1 0.5704 Order selected 1 sigma 2 estimated as 9431 > ar(Y, aic=FALSE,…

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Fundamentals of Differential Equations (9th Edition)

Ch. 4.1 - Verify that for b = 0 and Fext(t) = 0, equation...Ch. 4.1 - If Fext(t) = 0, equation (3) becomes my + by + ky...Ch. 4.1 - Show that if Fext(t) = 0, m = 1, k = 9, and b = 6,...Ch. 4.1 - Verify that y = sin 3t + 2 cos 3t is a solution to...Ch. 4.1 - Prob. 5E Ch. 4.1 - An external force F(t) = 2 cos 2t is applied to a...Ch. 4.1 - Prob. 7E Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - Prob. 10E

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The general solution to the differential equation 10 y ″ + y ′ − 3 y = t − e t 2 .

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