You need first solve the auxiliary equation of` y''' + y = 0` , such that:

`r^3 + 1 = 0`

You need to use the following formula, such that:

`a^3 + b^3 = (a + b)(a^2 - ab + b^2)`

Reasoning by analogy, yields:

`r^3 + 1 =...

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You need first solve the auxiliary equation of` y''' + y = 0` , such that:

`r^3 + 1 = 0`

You need to use the following formula, such that:

`a^3 + b^3 = (a + b)(a^2 - ab + b^2)`

Reasoning by analogy, yields:

`r^3 + 1 = (r + 1)(r^2 - r + 1)`

`r^3 + 1 = 0 => (r + 1)(r^2 - r + 1) = 0`

Using the zero product rule, yields:

`r + 1 = 0 => r = -1`

`r^2 - r + 1 = 0 => r_(1,2) = (1 +- sqrt(1 - 4))/2`

`r_(1,2) = (1 +-isqrt3)/2`

Since the roots of auxiliary equation are `r = -1` and `r_(1,2) = (1 +-isqrt3)/2` , the solution to the complementary equation is the following, such that:

`y_c = c_1*e^(-x) + e^(x/2)(c_2 cos ((sqrt3/2)x) + c_3 sin ((sqrt3/2)x))`

You need to find a particular solution, hence, you should try `y_p(x) = A*e^(2x)` , such that:

`y_p''' + y_p = e^(2x)`

Differentiating `y_p` with respect to `x` , yields:

`y'_p = 2A*e^(2x)`

`y''_p = 4A*e^(2x)`

`y'''_p = 8A*e^(2x)`

Replacing ` 8A*e^(2x)` for `y'''_p` and `A*e^(2x)` for `y_p` yields:

`8A*e^(2x) + A*e^(2x) = e^(2x)`

Factoring out `e^(2x)` yields:

`e^(2x)(8A + A) = e^(2x)`

Reducing duplicate factors yields:

`9A = 1 => A = 1/9`

Hence, evaluating the particular solution to the given equation, yields:

`y_p = (1/9)e^(2x)`

**Since the general solution of a nonhomogeneous differential equation can be written as `y = y_c + y_p` yields:**

`y(x) = c_1*e^(-x) + e^(x/2)(c_2 cos ((sqrt3/2)x) + c_3 sin ((sqrt3/2)x)) + (1/9)e^(2x)`