User:Egm4313.s12.team9.tjl

=R2.3: Find a general solution for problems 3 and 4 (K2011 pg. 59)=

Problem 3
Given: $$ y'' + 6y' + 8.96y = 0 $$ $$ \lambda^{2} + 6\lambda + 8.96 = 0 $$ Using the Quadratic Equation: $$ a=1, b=6, c=8.96 $$ $$ \frac{-b\frac{+}{-}\sqrt{b^{2} - 4ac}}{2a} $$ $$ \frac{-6\frac{+}{-}\sqrt{6^{2} - 35.84}}{2}$$ $$ \frac{-6 + 0.4}{2}$$ $$ \frac{-6 - 0.4}{2}$$ Root 1: -2.8 Root 2: -3.2 General Solution: $$ y = c_1e^{-2.8x} + c_2e^{-3.2x} $$

Problem 4
Given: $$ y'' + 4y' + (pi^{2}+4)y = 0 $$ $$ \lambda^{2} + 4\lambda + 13.87 = 0 $$ Using the Quadratic Equation: $$ a=1, b=4, c=13.87 $$ $$ \frac{-b\frac{+}{-}\sqrt{b^{2} - 4ac}}{2a} $$ $$ \frac{-4\frac{+}{-}\sqrt{4^{2} - 55.48}}{2}$$ $$ \frac{-4 + 6.28i}{2}$$ $$ \frac{-4 - 6.28i}{2}$$ Root 1: -2 + 3.14i Root 2: -2 - 3.14i General Solution: $$ y = c_1e^{-2x + 3.14ix} + c_2e^{-2x-3.14ix} $$

=R2.7: Develop the MacLaurin Series for $$ e^{t}, cos(t), and sin(t) $$= $$ P_n(t) = f(a) + f'(a)(t-a) + \frac{f''(a)}{2!}(t-a)^{2} +...+\frac{f^{n}(a)}{n!}(t-a)^{n} $$

MacLaurin Series(Taylor Series at t=0) for $$ e^{t} $$
$$ f(t) = e^{t}.....f(0)=1 $$ $$ f'(t) = e^{t}.....f'(0)=1 $$ $$ f(t) = e^{t}.....f(0)=1 $$ $$ f(t) = e^{t}.....f(0)=1 $$ $$ f'(t) = e^{t}.....f'(0)=1 $$ $$ P_5(t) = 1 + t + \frac{t^{2}}{2!} + \frac{t^{3}}{3!} + \frac{t^{4}}{4!} $$

MacLaurin Series(Taylor Series at t=0) for $$ cos(t) $$
$$ f(t) = \cos{t}.....f(0)=1 $$ $$ f'(t) = -\sin{t}.....f'(0)=0 $$ $$ f(t) = -\cos{t}.....f(0)=-1 $$ $$ f(t) = \sin{t}.....f(0)=0 $$ $$ f'(t) = \cos{t}.....f'(0)=1 $$ $$ P_5(t) = 1 - \frac{t^{2}}{2!} + \frac{t^{4}}{4!} $$

MacLaurin Series(Taylor Series at t=0) for $$ sin(t) $$
$$ f(t) = \sin{t}.....f(0)=0 $$ $$ f'(t) = \cos{t}.....f'(0)=1 $$ $$ f(t) = -\sin{t}.....f(0)=0 $$ $$ f(t) = -\cos{t}.....f(0)=-1 $$ $$ f'(t) = \sin{t}.....f'(0)=0 $$ $$ P_5(t) = t - \frac{t^{3}}{3!} $$