Solve the i.v.p for $y''+4y'+5y=0, y(\frac{\pi}{2})=1/2, y'(\frac{\pi}{2})=-2$
I solved using the quadratic form. and I got $\lambda = \frac{(-4 \pm 2i)}{2}$, which for $\lambda 1,2= 2+2i$.
And then for both my $\alpha$ and my $\beta$, I got 2.
$y(t)=Ae^{(2t)}\cos(2t)+Be^{(2t)}\sin(2t)$ bringing me to $y(t)=e^{(2t)}(A\cos(2t)+B\sin(2t))$
And then I would
(1) $y(t)=e^{(2(\pi/2))}(A\cos(2(\pi/2)+B\sin(2(\pi/2)))=1/2 = -e^{(\pi/2)}*(A+B)=1/2$
(2) $y'(t) =e^{(2(\pi/2))}(-A\sin(2(\pi/2)+B\cos(2(\pi/2)))=-2 = -e^{(\pi/2)}*(A+B)=-2$
I am not quite sure how to find my $A$ and $B$, but would my $A=1/2$ and my $B=-2$
Thus giving me $y(t)=1/2e^{(\pi-2t)} (\sin(t)+2\cos(t))$?