Question

y''+2y'+2y=2 y(0)=0,y'(0)=1

Answer 1

Characteristic equation

${k^2} + 2k + 2 = 0\\ D = 4 - 8 = - 4\\ {k_1} = \frac{{ - 2 - 2i}}{2} = - 1 - i\\ {k_2} = - 1 + i$

We obtain the general solution of the homogeneous equation

${y_0} = {e^{ - x}}\left( {{C_1}\cos x + {C_2}\sin x} \right)$

We will seek a particular solution of the equation in the form

$Y = A \Rightarrow Y' = Y'' = 0$

then

$0 + 0 + 2A = 2 \Rightarrow A = 1$

Then

$Y = A,\,\,y = {y_0} + Y = {e^{ - x}}\left( {{C_1}\cos x + {C_2}\sin x} \right) + 1$

$y' = - {e^{ - x}}\left( {{C_1}\cos x + {C_2}\sin x} \right) + {e^{ - x}}\left( { - {C_1}\sin x + {C_2}\cos x} \right)$

$y(0) = 0,\,y'(0) = 1 \Rightarrow \left\{ {\begin{matrix} {{C_1} +1= 0}\\ { - {C_1} + {C_2} = 1} \end{matrix}} \right. \Rightarrow {C_1} = -1,\,\,{C_2} = 0$

Answer: $y = - {e^{ - x}}\cos x + 1$