# Find the general solution of the ODE. {eq}y''' + y = 0{/eq}

## Question:

Find the general solution of the ODE.

{eq}y''' + y = 0{/eq}

## General solution of the ordinary differential equation:

If {eq}y = {y_1}, y = {y_2},...,y = {y_n} {/eq} are the linear independent solutions of {eq}\left( {{a_0}{D^n} + {a_1}{D^{n - 1}} + {a_2}{D^{n - 2}} + ... + {a_n}} \right)y = 0 {/eq} then {eq}y = {c_1}{y_1} + {c_2}{y_2} + ... + {c_n}{y_n} {/eq} is the general solution of the differential equation {eq}\left( {{a_0}{D^n} + {a_1}{D^{n - 1}} + {a_2}{D^{n - 2}} + ... + {a_n}} \right)y = 0 {/eq}, where {eq}{c_1},{c_2},....,{c_n} {/eq} are n arbitrary constants. The general solutions for real and different roots is {eq}y = {c_1}{e^{{m_1}x}} + {c_2}{e^{{m_2}x}} + {c_3}{e^{{m_3}x}} + ... + {c_n}{e^{{m_n}x}} {/eq} and for imaginary roots is {eq}y = {c_1}{e^{{m_1}x}} + {e^{\alpha x}}\left( {{c_2}\cos \beta x + {c_3}\sin \beta x} \right) {/eq}.

## Answer and Explanation: 1

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Given:

• The ordinary differential equation is given by {eq}y''' + y = 0 {/eq}.

The given equation can be re-written as:

{eq}{D^3}y + y =...

See full answer below.