User:Egm4313.s12.team3.katz/Brian Katz Report 2 Questions

Report 2
R2.1 (sec3 page 7) Find solution for 2 systems and plot R2.2 (sec5 page 7) L2 ODE no excitation R2.3 (sec5 page 7) similar to probs in previous report R2.4 (sec5 page 7) L2 find general solution R2.5 (sec5 page 7) Given solution find the ODE R2.6 (sec5 page 7) S-D-M system in seris find values

Problem Statement
R2.6 (sec5 page 7) S-D-M system in series find values

Problem Solution
Figure of a spring-dashpot-mass system in series Using the final equation describing the spring-dashpot-mass system:
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$$\displaystyle f(t)=m({y_k}''+\frac{k}{c}{y_k}')+ky_k \;\;\; becomes $$ $$\displaystyle f(t)=m{y_k}''+m\frac{k}{c}{y_k}'+ky_k $$ Now using the homogeneous equation:
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$$\displaystyle (\lambda-(-3)^2) = {\lambda}^2-6\lambda+9=0 $$ So replacing $$\lambda$$ with y':
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$$\displaystyle f(t)={y''}^2+6y'+9=0 $$ So now
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$$  \displaystyle m=1 $$ $$ \displaystyle m\frac{k}{c}=6 \;\;\; \rightarrow \;\;\; \frac{(1)(9)}{c}=6 \;\;\; \rightarrow  \;\;\; 9=6c \;\;\; \rightarrow \;\;\; c=1.5 $$ $$ \displaystyle k=9 $$
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Problem Statement
R2.7 (sec6 page 7) MacLaurin

Part A
You must find the first few non-zero derivative values to develop the Taylor series at t=0 (MacLaurin series):
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$$\displaystyle f(t) = e^t \,\,\,\,\,\,\,, \,\,\,\,\,\, f(0) = e^0 = 1 $$ $$\displaystyle f'(t) = e^t \,\,\,\,\,, \,\,\,\,\, f'(0) = e^0 = 1 $$ $$\displaystyle f(t) = e^t \,\,\,\,, \,\,\,\, f(0) = e^0 = 1 $$ $$\displaystyle f(t) = e^t \,\,\,, \,\,\, f(0) = e^0 = 1 $$
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The general form of Taylor series is:
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$$\displaystyle P_n(t)= f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots $$ For this problem the general form becomes:
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$$\displaystyle P_n(t)= f(0)+\frac {f'(0)}{1!} (t-0)+ \frac{f''(0)}{2!} (t-0)^2+\frac{f^{(3)}(0)}{3!}(t-0)^3+ \cdots $$ $$\displaystyle P_n(t)= 1+t+\frac {1}{2} (t)^2+ \frac{1}{6} (t)^3+ \cdots $$ In Sigma notation:
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$$\displaystyle e^t=\sum_{n=0}^{\infty }\frac{t^n}{n!} $$
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Part B
You must find the first few non-zero derivative values to develop the Taylor series at t=0 (MacLaurin series):
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$$\displaystyle f(t) = cos(t) \,\,\,\,\,\,\,\,\,\,\,\,\,\,, \,\,\,\,\,\,\,\,\, f(0) = cos(0) = 1 $$ $$\displaystyle f'(t) = -sin(t) \,\,\,\,\,\,\,\,, \,\,\,\,\,\,\,\, f'(0) = -sin(0) = 0 $$ $$\displaystyle f(t) = -cos(t) \,\,\,\,\,\,\,, \,\,\,\,\,\,\, f(0) = -cos(0) = -1 $$ $$\displaystyle f(t) = sin(t) \,\,\,\,\,\,\,\,\,\,\,, \,\,\,\,\, f(0) = sin(0) = 0 $$ $$\displaystyle f^{(4)}(t) = cos(t) \,\,\,\,\,\,\,\,\,, \,\,\,\, f^{(4)}(0) = cos(0) = 1 $$ $$\displaystyle f^{(5)}(t) = -sin(t) \,\,\,\,, \,\,\,\, f^{(5)}(0) = -sin(0) = 0 $$ $$\displaystyle f^{(6)}(t) = -cos(t) \,\,\,\,, \,\,\,\, f^{(6)}(0) = -cos(0) = -1 $$
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The general form of Taylor series is:
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$$\displaystyle P_n(t)= f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots $$ For this problem the general form becomes:
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$$\displaystyle P_n(t)= f(0)+ \frac{f''(0)}{2!} (t-0)^2+\frac{f^{(4)}(0)}{4!}(t-0)^4+\frac{f^{(6)}(0)}{6!}(t-0)^6\cdots $$ $$\displaystyle P_n(t)= 1-\frac{1}{2}(t)^2+ \frac{1}{24} (t)^4- \frac{1}{720} (t)^6 +\cdots $$ In Sigma notation:
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$$\displaystyle cos(t)=\sum_{n=0}^{\infty }\frac{(-1)^kt^{2k}}{(2k)!} $$
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Part C
You must find the first few non-zero derivative values to develop the Taylor series at t=0 (MacLaurin series):
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$$\displaystyle f(t) = sin(t) \,\,\,\,\,\,\,\,\,\,\,\,\,\,, \,\,\,\,\,\, f'''(0) = sin(0) = 0 $$ $$\displaystyle f'(t) = cos(t) \,\,\,\,\,\,\,\,\,\,\,\,\,, \,\,\,\,\,\,\,\, f'(0) = cos(0) = 1 $$ $$\displaystyle f(t) = -sin(t) \,\,\,\,\,\,\,, \,\,\,\,\,\,\, f(0) = -sin(0) = 0 $$ $$\displaystyle f(t) = -cos(t) \,\,\,\,\,\,, \,\,\,\,\,\, f(0) = -cos(0) = -1 $$ $$\displaystyle f^{(4)}(t) = sin(t) \,\,\,\,\,\,\,\,\,, \,\,\,\, f^{(4)}(0) = sin(0) = 0 $$ $$\displaystyle f^{(5)}(t) = cos(t) \,\,\,\,\,\,\,\,\,, \,\,\,\, f^{(5)}(0) = cos(0) = 1 $$ $$\displaystyle f^{(6)}(t) = -sin(t) \,\,\,\,, \,\,\,\, f^{(6)}(0) = -sin(0) = 0 $$ $$\displaystyle f^{(7)}(t) = -cos(t) \,\,\,\,, \,\,\,\, f^{(7)}(0) = -cos(0) = -1 $$
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The general form of Taylor series is:
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$$\displaystyle P_n(t)= f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots $$
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For this problem the general form becomes:
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$$\displaystyle P_n(t)= \frac {f'(0)}{1!} (t-0)+\frac{f^{(3)}(0)}{3!}(t-0)^3+ \frac{f^{(5)}(0)}{5!}(t-0)^5+ \frac{f^{(7)}(0)}{7!}(t-0)^7 +\cdots $$ $$\displaystyle P_n(t)= t-\frac{1}{6}(t)^3+ \frac{1}{120} (t)^5- \frac{1}{5040} (t)^7 +\cdots $$ In Sigma notation:
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$$\displaystyle sin(t)=\sum_{n=0}^{\infty }\frac{(-1)^kt^{2k+1}}{(2k+1)!} $$
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