Question

Find the general solution of the following differential equation
3y" +2y' +y = 0

Answer 1

Corresponding (auxiliary) equation

$$3r^2+2r+1=0$$

$$D=(2)^2-4(3)(1)=-8$$

$$r=\dfrac{-2\pm\sqrt{-8}}{2(3)}=\dfrac{-1\pm\sqrt{2}}{3}$$

$$r_1=\dfrac{-1-\sqrt{2}}{3},r_2=\dfrac{-1+\sqrt{2}}{3}$$

The general solution of the given differential equation is

$$y(t)=c_1e^{-{1 \over 3}t}\cos({\sqrt{2} \over 3}t)+c_2e^{-{1 \over 3}t}\sin({\sqrt{2} \over 3}t)$$