To solve the second order differential equation 3y” + 2y’ + y = 0, we will use the characteristic equation method.
First, we rewrite the equation in standard form, dividing through by 3:
y” + (2/3)y’ + (1/3)y = 0
The associated characteristic equation for this differential equation is obtained by substituting y = e^{rt}, where r is a constant:
r^2 + (2/3)r + (1/3) = 0
Next, we will apply the quadratic formula to find the roots:
r = rac{-b ext{±} ext{sqrt}(b^2 – 4ac)}{2a}
Here, a = 1, b = 2/3, and c = 1/3. Plugging in these values, we get:
r = rac{- rac{2}{3} ext{±} ext{sqrt}(( rac{2}{3})^2 – 4 imes 1 imes rac{1}{3})}{2 imes 1}
r = rac{- rac{2}{3} ext{±} ext{sqrt}( rac{4}{9} – rac{4}{3})}{2}
r = rac{- rac{2}{3} ext{±} ext{sqrt}( rac{4}{9} – rac{12}{9})}{2}
r = rac{- rac{2}{3} ext{±} ext{sqrt}(- rac{8}{9})}{2}
We simplify the square root of the negative value:
r = rac{- rac{2}{3} ext{±} rac{2i ext{sqrt}(2)}{3}}{2}
r = - rac{1}{3} ext{±} rac{i ext{sqrt}(2)}{3}
This gives us a pair of complex conjugate roots:
r_1 = - rac{1}{3} + rac{i ext{sqrt}(2)}{3}, r_2 = - rac{1}{3} – rac{i ext{sqrt}(2)}{3}
For complex roots of the form r = eta ± i heta, the general solution of the differential equation can be given by:
y(t) = e^{eta t}(C_1 ext{cos}( heta t) + C_2 ext{sin}( heta t))
In our case, eta = - rac{1}{3} and heta = rac{ ext{sqrt}(2)}{3}.
Substituting into the general solution form, we have:
y(t) = e^{- rac{1}{3} t}(C_1 ext{cos}( rac{ ext{sqrt}(2)}{3} t) + C_2 ext{sin}( rac{ ext{sqrt}(2)}{3} t))
Thus, the general solution to the differential equation 3y” + 2y’ + y = 0 is:
y(t) = e^{- rac{1}{3} t}(C_1 ext{cos}( rac{ ext{sqrt}(2)}{3} t) + C_2 ext{sin}( rac{ ext{sqrt}(2)}{3} t))