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# 285hw7 - HW 71 1 Sec 3.6 3 We have x00 100x = 225 cos 5t...

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HW 7 1 1. Sec. 3.6: 3. We have x 00 + 100 x = 225 cos 5 t + 300 sin 5 t; x (0) = 375 ; x 0 (0) = 0 : The characteristic equation is r 2 + 100 = 0 = ) r = ° 10 i: The compli- mentary solution is x c ( t ) = c 1 cos 10 t + c 2 sin 10 t: r = 5 i is not a root of the characteristic equation. Hence, the particular solution is in the form: x p ( t ) = A cos 5 t + B sin 5 t; then x 0 p ( t ) = ± 5 A sin 5 t + 5 B cos 5 t; and x 00 p ( t ) = ± 25 A cos 5 t ± 25 B sin 5 t: Now we plug x p ( t ) and x 00 p ( t ) into the given equation. We get ± 25 A cos 5 t ± 25 B sin 5 t + 100 A cos 5 t + 100 B sin 5 t = 225 cos 5 t + 300 sin 5 t: Thus we have the following system ± 25 A + 100 A = 225 ; ± 25 B + 100 B = 300 : The solution of the system is A = 3 ; B = 4 : So the particular solution is x p ( t ) = 3 cos 5 t + 4 sin 5 t: The general solution is x ( t ) = x c ( t ) + x p ( t ) = c 1 cos 10 t + c 2 sin 10 t + 3 cos 5 t + 4 sin 5 t . 1 Note to the grader: Please grade the following problems: Sec. 3.6:7, Sec. 3.6:15, Sec. 3.6: 23, Sec. 3.6:25, Sec. 3.6:27 (20pts for each problem). Please note that students do not have to construct sketches. 1

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Then x 0 ( t ) = ± 10 c 1 sin 10 t + 10 c 2 cos 10 t ± 15 sin 5 t + 20 cos 5 t: Now using the initial conditions we get the following system x (0) = c 1 + 3 = 375 ; x 0 (0) = 10 c 2 + 20 = 0 : The solution of the system is c 1 = 373 ; c 2 = ± 2 : Thus we have x ( t ) = 372 cos 10 t ± 2 sin 10 t + 3 cos 5 t + 4 sin 5 t: Now we want to express the solution of the given initial value problem as a sum of two oscillations: q 372 2 + ( ± 2) 2 = p 138 388; q (3) 2 + (4) 2 = 5 : 372 cos 10 t ± 2 sin 10 t = p 138 388 ° 372 p 138 388 cos 10
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