285hw7 - HW 7 1 1. Sec. 3.6: 3. We have x 00 + 100 x = 225...

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Unformatted text preview: 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 : 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 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 Then x ( 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) = 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...
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285hw7 - HW 7 1 1. Sec. 3.6: 3. We have x 00 + 100 x = 225...

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