# Differentiate to get y p k sin t m cos t y p k cos t

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Differentiate to get y ± p = K sin t + M cos t y ±± p = K cos t M sin t . Substitute this into the given ODE. Abbreviate C = cos t , S = sin t . We obtain ( KC MS ) + 3( KS + MC ) + 3 . 25( KC + MS ) = 3 C 1 . 5 S . Collect the C -terms and equate their sum to 3 K + 3 M + 3 . 25 K = 3 so that 2 . 25 K + 3 M = 3 . Collect the S -terms and equate their sum to 1 . 5: M 3 K + 3 . 25 M = 15 so that 3 K + 2 . 25 M = 1 . 5 . We solve the equation for the C -terms for M and get M = 1 0 . 75 K . We substitute this into the previous equation (equation for the S -terms), simplify, and get a value for K 2 . 25(1 0 . 75 K ) 3 K = 1 . 5 thus K = 0 . 8 . Hence M = 1 0 . 75 K = 1 0 . 75 · 0 . 8 = 0 . 4 . This con fi rms the answer on p. A8 that the transient solution is y = y h + y p = e 1 . 5 t ( A cos t + B sin t ) + 0 . 8 cos t + 0 . 4 sin t . 17. Initial value problem. The homogeneous ODE is y ±± + 4 y = 0. Its characteristic equation is λ 2 + 4 = ( λ 2)( λ + 2) = 0 . It has the roots 2 and 2, so that a general solution of the homogeneous ODE is y h = c 1 e 2 t + c 2 e 2 t . Next we need a particular solution y p of the given ODE. The right-hand side of that ODE is sin t + 1 3 sin 3 t + 1 5 cos 5 t . Using the method of undetermined coef fi cients (Sec. 2.7), set y p = y p 1 + y p 2 + y p 3 .
Chap. 2 Second-Order Linear ODEs 29 For the fi rst part of y p y p 1 = K 1 sin t + M 1 cos t . Differentiation (chain rule) then gives y ± p 1 = K 1 cos t M 1 sin t , y ±± p 1 = K 1 sin t M 1 cos t . Similarily for y p 2 : y p 2 = K 2 sin 3 t + M 2 cos 3 t . y ± p 2 = 3 K 2 cos 3 t 3 M 2 sin 3 t , y ±± p 2 = 9 K 2 sin 3 t 9 M 2 cos 3 t . Finally, for y p 3 , we have y p 3 = K 3 sin 5 t + M 3 cos 5 t . y ± p 3 = 5 K 3 cos 5 t 5 M 3 sin 5 t , y ±± p 3 = 25 K 3 sin 5 t 25 M 3 cos 5 t . Denote S = sin t , C = cos t ; S = sin 3 t , C = cos 3 t ; S ∗∗ = sin 5 t , C ∗∗ = cos 5 t . Substitute y p and y ±± p = y ±± p 1 + y ±± p 2 + y ±± p 3 (with the notation S , C , S , C , S ∗∗ , and C ∗∗ ) into the given ODE y ±± + 4 y = sin t + 1 3 sin 3 t + 1 5 cos 5 t and get a very long equation, stretching over two lines: ( K 1 S M 1 C 9 K 2 S 9 M 2 C 25 K 3 S ∗∗ 25 M 3 C ∗∗ ) + (4 K 1 S + 4 M 1 C + 4 K 2 S + 4 M 2 C + 4 K 3 S ∗∗ + 4 M 3 C ∗∗ ) = S + 1 3 S + 1 5 C ∗∗ . Collect the S -terms and equate the sum of their coef fi cients to 1 because the right-hand side of the ODE has one term sin t , which we denoted by S . [ S -terms] K 1 + 4 K 1 = 1 so that K 1 = 1 3 . Simlarly for C -terms [ C -terms] M 1 + 4 M 1 = 0 so that M 1 = 0 . Then for S -terms, C -terms [ S -terms] 9 K 2 + 4 K 2 = 1 3 so that K 2 = 1 15 [ C -terms] 9 M 2 + 4 M 2 = 0 so that M 2 = 0 . And fi nally for S ∗∗ -terms, C ∗∗ -terms [ S ∗∗ -terms] 25 K 3 + 4 K 3 = 1 5 so that K 3 = 1 105 [ C ∗∗ -terms] 25 M 3 + 4 M 3 = 0 so that M 3 = 0 . Hence the general solution of the given ODE is y ( t ) = c 1 e 2 t + c 2 e 2 t + 1 3 sin t 1 15 sin 3 t 1 105 sin 5 t .
30 Ordinary Differential Equations (ODEs) Part A We now consider the initial conditions. The fi rst condition y (0) = 0 gives y (0) = c 1 + c 2 + 0 = 0, so that c 1 = c 2 . For the second condition y ± (0) = 3 35 , we need to compute y ±