Find the general solution of s'' + 2s' + 2s = 2t^3 - 2

Find the general solution of $s'' + 2s' + 2s = 2t^3 - 2$.

Step 1: Homogeneous Solution

\[ r^2+2r+2=0 \implies r=\frac{-2\pm\sqrt{4-8}}{2}=-1\pm i. \]

\[ S_h = e^{-t}\left(C_1\cos t + C_2\sin t\right). \]

Step 2: Particular Solution

The right-hand side $f(t)=2t^3-2$ is a polynomial of degree 3, and $r=0$ is not a root of the characteristic equation, so the multiplicity is $s=0$ and we try $s^*=At^3+Bt^2+Ct+D$:

\[ s^{*\prime}=3At^2+2Bt+C, \qquad s^{*\prime\prime}=6At+2B. \]

Substitute into the equation: $(6At+2B)+2(3At^2+2Bt+C)+2(At^3+Bt^2+Ct+D)=2t^3-2$, $2At^3+(6A+2B)t^2+(6A+4B+2C)t+(2B+2C+2D)=2t^3-2$.

Step 3: Match Coefficients

\[ \begin{aligned} t^3:&\ 2A=2 \implies A=1, \\ t^2:&\ 6A+2B=0 \implies 6+2B=0 \implies B=-3, \\ t^1:&\ 6A+4B+2C=0 \implies 6-12+2C=0 \implies C=3, \\ t^0:&\ 2B+2C+2D=-2 \implies -6+6+2D=-2 \implies D=-1. \end{aligned} \] So $s^*=t^3-3t^2+3t-1$, which is recognizable as the perfect cube:

\[ s^* = (t-1)^3. \]

Step 4: Verification

\[ s^{*\prime}=3(t-1)^2, \qquad s^{*\prime\prime}=6(t-1). \]

\[ s^{*\prime\prime}+2s^{*\prime}+2s^* = 6(t-1)+6(t-1)^2+2(t-1)^3 = 2t^3-2.\ \checkmark \]

Step 5: General Solution