Final answer:
The general solution of the differential equation 9y" + 6y' + y = 0 is y = c₁e^(-t/3) + c₂te^(-t/3), where c₁ and c₂ are arbitrary constants.
Step-by-step explanation:
General Solution of the Differential Equation
To find the general solution of the differential equation 9y" + 6y' + y = 0, we can assume a solution of the form y = e^(rt), where r is a constant. Differentiating y twice gives us y" = r^2e^(rt) and y' = re^(rt). Substituting these expressions into the differential equation gives us 9r^2e^(rt) + 6re^(rt) + e^(rt) = 0.
Dividing through by e^(rt), we get the characteristic equation: 9r^2 + 6r + 1 = 0. Solving this quadratic equation for r gives us two distinct roots: r₁ = -1/3 and r₂ = -1/3. Therefore, the general solution of the differential equation is y = c₁e^(-t/3) + c₂te^(-t/3), where c₁ and c₂ are arbitrary constants.