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

3r² + 2r + 1 = 0

The formula for the standard form of quadratic equation ax² + bx + c = 0 is written as

x = [−b ± √(b² − 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) ].

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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) ].