Author Topic: tut0402 quiz4  (Read 398 times)

yuhan cheng

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tut0402 quiz4
« 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: $y=\sqrt{2} e^{-t+\pi / 4} \cos t+\sqrt{2} e^{-t+\pi / 4} \sin t$