### Author Topic: TUT0402 Quiz4  (Read 3976 times)

#### Jingjing Cui

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##### TUT0402 Quiz4
« on: October 18, 2019, 01:59:46 PM »
Given
$$y"(t)+2y'(t)+2=0\\ y(\frac{\pi}{4})=2 \;\;\; y'(\frac{\pi}{4})=-2\\$$
Solution:
$$r^2+2r+2=0\\ r_1=\frac{-b+(b^2-4ac)^{1/2}}{2a}\\ r_2=\frac{-b-(b^2-4ac)^{1/2}}{2a}\\ r_1=\frac{-2+(4-4*2)^{1/2}}{2}=-1+i\\ r_2=\frac{-2-(4-4*2)^{1/2}}{2}=-1-i\\ \lambda=-1 \; \mu=1\\ y(t)=c_1e^{-t}cos(t)+c_2e^{-t}sin(t)\\$$
Substituting the initial conditions:
$$\\ 2=c_1e^{-\frac{\pi}{4}}cos(\frac{\pi}{4})+c_2e^{-\frac{\pi}{4}}sin(\frac{\pi}{4})\\ 2=c_1e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}+c_2e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}\\ 2\sqrt2=c_1e^{-\frac{\pi}{4}}+c_2e^{-\frac{\pi}{4}} \\ y'(t)=-c_1e^{-t}cos(t)-c_1e^{-t}sin(t)-c_2e^{-t}sin(t)+c_2e^{-t}cos(t)\\ \\ -2=-c_1e^{-\frac{\pi}{4}}cos(\frac{\pi}{4})-c_1e^{-\frac{\pi}{4}}sin(\frac{\pi}{4})-c_2e^{-\frac{\pi}{4}}sin(\frac{\pi}{4})+c_2e^{-\frac{\pi}{4}}cos(\frac{\pi}{4})\\ -2=-c_1e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}-c_1e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}-c_2e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}+c_2e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}\\ -2=-2c_1e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}\\ c_1=e^{\frac{\pi}{4}}\sqrt2\\ 2\sqrt2=\sqrt2e^{\frac{\pi}{4}}e^{-\frac{\pi}{4}}+c_2e^{-\frac{\pi}{4}}\\ 2\sqrt2=\sqrt2+c_2e^{-\frac{\pi}{4}}\\ c_2=e^{\frac{\pi}{4}}\sqrt2\\ y(t)=\sqrt2e^{\frac{\pi}{4}}e^{-t}cos(t)+\sqrt2e^{\frac{\pi}{4}}e^{-t}sin(t)\\$$