How can I solve this initial value problem?

$$ y'' + 2y' + 2y = 0,$$ given $y\,(\pi/4)=2$ and $y'(\pi/4)=0$.

I've found $y(t)=e^{-t} \left(C_1\cos t + C_2\sin t \right)$ but I wasn't able to find $C_1$ and $C_2$. How can I find them?


Using the $y(t)$ you found (which is correct), substitute $\dfrac{\pi}{4}$ into $y(t)$, yielding:

$$\dfrac{1}{\sqrt{2}}e^{-\pi/4}(c_1 + c_2) = 2$$

Substitute $\dfrac{\pi}{4}$ into $y'(t)$, yielding:

$$-\sqrt{2}e^{-\pi/4} c_1 = 0$$

This gives us $c_1 = 0$.

Can you now find $c_2$?

  • $\begingroup$ Thank you so much for the answer! $\endgroup$
    – cecemelly
    Nov 9 '13 at 19:32
  • 1
    $\begingroup$ You are quite welcome. If $c_1$ was not equal to zero, we would be solving a $2 x 2$ system, which is typically what happens. Regards $\endgroup$
    – Amzoti
    Nov 9 '13 at 19:35
  • $\begingroup$ Well done! (And great feedback, too!) $\endgroup$
    – amWhy
    Nov 10 '13 at 3:01
  • $\begingroup$ I agree with Amy. @Amzoti so :+) $\endgroup$
    – Mikasa
    Nov 20 '13 at 14:21

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