# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-4 => Topic started by: yuhan cheng on October 18, 2019, 02:01:58 PM

Title: tut0402 quiz4
Post by: yuhan cheng on October 18, 2019, 02:01:58 PM
22.$$y^{\prime \prime}+2 y^{\prime}+2 y=0, y(\pi / 4)=2, y^{\prime}(\pi / 4)=-2$$
$$\begin{array}{c}{y^{\prime \prime}+2 y^{\prime}+2 y=0} \\ {r^{2}+2 r+2=0} \\ {r=-1 \pm i} \\ {y=c_{1} e^{-t} \cos t+c_{2} e^{-t} \sin t}\end{array}$$

We know that general solution is $y=c_{1} e^{-t} \cos t+c_{2} e^{-t} \sin t$

Initial condition $y(\pi / 4)=2$
$$\begin{array}{l}{c_{1} e^{-\pi / 4} \cdot \frac{\sqrt{2}}{2}+c_{2} e^{-\pi / 4} \cdot \frac{\sqrt{2}}{2}=2} \\ {c_{1}+c_{2}=2 \sqrt{2} e^{\pi / 4}}\end{array}$$
$$\begin{array}{l}{y^{\prime}=-c_{1} e^{-t} \cos t-c_{1} e^{-t} \sin t-c_{2} e^{-t} \sin t+c_{2} e^{-t} \cos t} \\ {y^{\prime}(\pi / 4)=-2} \\ {-c_{1} e^{-\pi / 4} \frac{\sqrt{2}}{2}-c_{1} e^{-\pi / 4} \frac{\sqrt{2}}{2}-c_{2} e^{-\pi / 4} \frac{\sqrt{2}}{2}+c_{2} e^{-\pi / 4} \frac{\sqrt{2}}{2}=-2} \\ {-\sqrt{2} c_{1}=-2 e^{\pi / 4}} \\ {c_{1}=\sqrt{2} e^{\pi / 4}} \\ {c_{2}=2 \sqrt{2} e^{\pi / 4}-\sqrt{2} e^{\pi / 4}=\sqrt{2} e^{\pi / 4}}\end{array}$$

Solution: $y=\sqrt{2} e^{-t+\pi / 4} \cos t+\sqrt{2} e^{-t+\pi / 4} \sin t$