The general solution to the equation

y+4y+5y\displaystyle {y}{''}+{4}{y}'+{5}{y} = 0

has the form eat(c1cos(bt)+c2sin(bt))\displaystyle {e}^{{{a}{t}}}{\left({c}_{{1}}{\cos{{\left({b}{t}\right)}}}+{c}_{{2}}{\sin{{\left({b}{t}\right)}}}\right)} where b0\displaystyle {b}\ge{0}.

a\displaystyle {a} =  

b\displaystyle {b} =  

Now suppose that the initial conditions are y(0)=1,y(0)=0\displaystyle {y}{\left({0}\right)}={1},\quad{y}'{\left({0}\right)}={0}. Then solve for:

c1\displaystyle {c}_{{1}} =  

c2\displaystyle {c}_{{2}} =