Currently I'm working on the following exercise:
$$ 4y''+ 4y'+ 5y = 0\,,\qquad y\left(\pi\right) = 1 \,,\quad y'\left(\pi\right) = 0 $$
My solution to this exercise is $$ {\rm e}^{-x/2}\left[\,{-\cos\left(x\right)/{\rm e}^{-\pi/2} - \sin\left(x\right)/{\rm e}^{-\pi/2}}\,\,\right], $$ but I'm not sure about this answer since Symbolab (https://www.symbolab.com/solver/simplify-calculator/4y''%2B4y'%2B5y%3D0%2C%20y%5Cleft(%5Cpi%5Cright)%3D1%2C%20y'%5Cleft(%5Cpi%20%5Cright)%3D0) tells me something else.
Can someone either correct my answer or confirm the correct answer?
Thanks!
The general solution is given by
$$y(x) = c_1 e^{-x/2} \sin(x) + c_2 e^{-x/2} \cos(x)$$
The derivative is given by
$$y'(x) = \dfrac{1}{2} e^{-x/2} (c_1 (2 \cos(x) - \sin(x)) - c_2 (2 \sin(x) + \cos(x)))$$
Evaluating $y(x)$ at $x = \pi$
$$y(\pi) = -c_2 e^{-\pi/2} = 1$$
Evaluating $y'(x)$ at $x = \pi$
$$y'(\pi) = \dfrac{1}{2} e^{-\pi/2} (c_2 - 2 c_1) = 0$$
Can you solve those two for $c_1$ and $c_2$?