Author Topic: TUT0402 Quiz4  (Read 316 times)

Jingjing Cui

  • Full Member
  • ***
  • Posts: 16
  • Karma: 5
    • View Profile
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:
$$
\\
2=c_1e^{-\frac{\pi}{4}}cos(\frac{\pi}{4})+c_2e^{-\frac{\pi}{4}}sin(\frac{\pi}{4})\\
2=c_1e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}+c_2e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}\\
2\sqrt2=c_1e^{-\frac{\pi}{4}}+c_2e^{-\frac{\pi}{4}}
\\
y'(t)=-c_1e^{-t}cos(t)-c_1e^{-t}sin(t)-c_2e^{-t}sin(t)+c_2e^{-t}cos(t)\\
\\
-2=-c_1e^{-\frac{\pi}{4}}cos(\frac{\pi}{4})-c_1e^{-\frac{\pi}{4}}sin(\frac{\pi}{4})-c_2e^{-\frac{\pi}{4}}sin(\frac{\pi}{4})+c_2e^{-\frac{\pi}{4}}cos(\frac{\pi}{4})\\
-2=-c_1e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}-c_1e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}-c_2e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}+c_2e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}\\
-2=-2c_1e^{-\frac{\pi}{4}}\frac{1}{\sqrt2}\\
c_1=e^{\frac{\pi}{4}}\sqrt2\\
2\sqrt2=\sqrt2e^{\frac{\pi}{4}}e^{-\frac{\pi}{4}}+c_2e^{-\frac{\pi}{4}}\\
2\sqrt2=\sqrt2+c_2e^{-\frac{\pi}{4}}\\
c_2=e^{\frac{\pi}{4}}\sqrt2\\
y(t)=\sqrt2e^{\frac{\pi}{4}}e^{-t}cos(t)+\sqrt2e^{\frac{\pi}{4}}e^{-t}sin(t)\\
$$