# Find the general solution of the given second-order differential equation. 3y'' + 2y' + y = 0

**Solution:**

Given that:

3y'' + 2y' + y = 0

The characteristic equation can be written as

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

The formula for the standard form of quadratic equation ax^{2} + bx + c = 0 is written as

x = [−b ± √(b^{2} − 4ac)] / 2a

From the given equation we know that

a = 3

b = 2

c = 1

Substituting it in the formula

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

x = [-2 ± √4 − 12] / 6

By further calculation

x = [-2 ± √ −8] / 6

x = [-2 ± √ −8] / 6

So we get

x = (-1/3) ± i (√ 2/3) [Since, √ −8 = 2i√ 2] …. (1)

The general solution of a second order with complex roots v ± wi is written as

y = evx [C cos(wx) + iD sin(wx)] …. (2)

From the equation (1), substitute v = -1/3 and w = √ 2/3 in (2)

y = e^{-x/3} [C cos(x√ 2/3) + iD sin(x√ 2/3) ]

Therefore, the general solution is y = e^{-x/3} [C cos(x√ 2/3) + iD sin(x√ 2/3) ].

## Find the general solution of the given second-order differential equation.

3y'' + 2y' + y = 0

**Summary: **

The general solution of the given second-order differential equation 3y'' + 2y' + y = 0 is y = e^{-x/3} [C cos(x√ 2/3) + iD sin(x√ 2/3) ].

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