Find the general solution of y'' + 2y' + 3y = 0

Find the general solution of $y'' + 2y' + 3y = 0$.

Step 1: Characteristic equation

Substituting $y=e^{rx}$: $r^2+2r+3=0.$

The discriminant is $\Delta=2^2-4(1)(3)=4-12=-8<0$, so the roots are complex conjugates: \[r=\frac{-2\pm\sqrt{-8}}{2}=\frac{-2\pm2i\sqrt2}{2}=-1\pm i\sqrt2 \implies \alpha=-1,\ \ \beta=\sqrt2.\]

Step 2: Fundamental system

Complex conjugate roots $r=\alpha\pm i\beta$ give the real fundamental system $e^{\alpha x}\cos\beta x,\ e^{\alpha x}\sin\beta x$: $y_1=e^{-x}\cos(\sqrt2\,x), \qquad y_2=e^{-x}\sin(\sqrt2\,x).$

Step 3: Wronskian

\[ W(y_1,y_2)=\begin{vmatrix}e^{-x}\cos(\sqrt2 x) & e^{-x}\sin(\sqrt2 x)\\ -e^{-x}\!\left(\cos(\sqrt2 x)+\sqrt2\sin(\sqrt2 x)\right) & e^{-x}\!\left(\sqrt2\cos(\sqrt2 x)-\sin(\sqrt2 x)\right)\end{vmatrix} = \sqrt2\,e^{-2x} \neq 0, \]

confirming $y_1,y_2$ are linearly independent for all $x$.

Step 4: General solution

The solution is recorded in the result below.