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**Quiz-4 / TUT0602 QUIZ 4**

« **on:**October 18, 2019, 03:30:57 PM »

Find the general solution of the given differential equation.

$$ y'' + 2y' + 2y = 0$$

Solve the equation $r^2+2r+2=0$ for $r$

$$ r = \frac{-2 \pm \sqrt{2^2-4(1)(2)} }{2(1)} $$

$$ = \frac{-2 \pm \sqrt{-4}}{2} = \frac {-2 \pm 2i}{2} = -1 \pm i$$

We know that $y = c_1e^{\lambda t}cos(\mu t) + c_2e^{\lambda t}sin(\mu t)$ and $\lambda = -1$ and $\mu = 1$

Therefore, the required general solution is

$$ y = c_1e^{-t}cos(t) + c_2e^{-t}sin(t) $$

$$ y'' + 2y' + 2y = 0$$

Solve the equation $r^2+2r+2=0$ for $r$

$$ r = \frac{-2 \pm \sqrt{2^2-4(1)(2)} }{2(1)} $$

$$ = \frac{-2 \pm \sqrt{-4}}{2} = \frac {-2 \pm 2i}{2} = -1 \pm i$$

We know that $y = c_1e^{\lambda t}cos(\mu t) + c_2e^{\lambda t}sin(\mu t)$ and $\lambda = -1$ and $\mu = 1$

Therefore, the required general solution is

$$ y = c_1e^{-t}cos(t) + c_2e^{-t}sin(t) $$