# Find the General Solution of the given Second-Order Differential Equation 3y'' + 2y' + y = 0

We will be using the concept of second-order differential equations to solve this.

## Answer: The General Solution of the given Second-Order Differential Equation 3y'' + 2y' + y = 0 is y = e^{-x/3} [Ccos(x√ 2/3) + iDsin(x√ 2/3)]

Let's solve this step by step.

**Explanation:**

Given: 3y'' + 2y' + y = 0

The characteristic equation for the given equation 3y'' + 2y' + y = 0 is:

3r^{2} + 2r + 1 = 0

Use the quadratic equation formula,

For a general quadratic equation ax^{2} + bx + c = 0,

x = [−b ± √(b^{2 }− 4ac)] / 2a

For the equation 3r^{2} + 2r + 1 = 0

a = 3, b = 2, c = 1.

x = [−2 ± √{(2)^{2 }− 4 × 3 × 1}] / 2(3)

x = [-2 ± √4^{ }− 12] / 6

x = [-2 ± √^{ }−8] / 6

x = [-2 ± √^{ }−8] / 6

x = (-1/3) ± i(√^{ }2/3) [Since, √^{ }−8 = 2i√^{ }2] ------------- (1)

We know that general solution of a second order differential equations with complex roots v ± wi is given by:

y = e^{vx} [Ccos(wx) + iDsin(wx)] -------------- (2)

From (1)

Substitute v = -1/3 and w = √^{ }2/3 in (2)

⇒ y = e^{-x/3} [Ccos(x√^{ }2/3) + iDsin(x√^{ }2/3) ]