Solving a linear, 2nd order ODE IVP two different ways - one does not lead to solutions

Question

So I've been looking at solving linear, 2nd order ODEs of the form $$ ay''+by'+cy=0$$ Now, if I solve the ODE $$ y''+4y'+5y=0$$ I should be able to express the general solution either as $$ y(t) = C_1 e^{(-2+i)t} + C_2 e^{(-2-i)t} $$ or, alternatively $$ y(t) = e^{-2t}(c_1 \cos( t) + c_2 sin( t))$$ With $c_1 = C_1 + C_2$ and $c_2 = i(C_1 - C_2)$. Both formulations should, at least in theory, be equivalent since I simply derived the second from the first using Euler's formula. However, for the initial conditions $y(0)=2$ and $y'(0)=-5$, if one tries to solve for $C_1$ and $C_2$ using the first equation one gets a contradiction. On the other hand, solving for $c_1$ and $c_2$ using the second formulation gives the perfectly happy values of $c_1 = 2$ and $c_2 = -1$. What's going on here?

Answer 1

$$y(t) = A e^{(-2+i)t} + Be^{(-2-i)t}$$ $$y'(t) = A(-2+i) e^{(-2+i)t} + B(-2-i)e^{(-2-i)t}$$ The initial conditions give ; $$(A,B)=(1+\dfrac i 2,1-\dfrac i2)$$ For the second solution: $$y(t) = e^{-2t}(c_1 \cos( t) + c_2 \sin( t))$$ $$y'(t) = -2e^{-2t}(c_1 \cos( t) + c_2 \sin( t))+ e^{-2t}(-c_1 \sin( t) + c_2 \cos( t))$$ The initial conditions give is: $$(c_1,c_2)=(2,-1)$$ So that you have indeed: $$c_1=A+B=2 \text { and } c_2=i(A-B)=i(i)=-1$$