Find the general solution of y''' + y = 0

Find the general solution of $y''' + y = 0$.

Step 1: Characteristic equation

Substituting $y=e^{rx}$ gives

\[ r^3+1=0. \]

This is a sum of cubes, $r^3+1^3$, which factors as

\[ (r+1)(r^2-r+1)=0. \]

Step 2: Solve for the roots

The linear factor gives $r_1=-1$. The quadratic factor $r^2-r+1=0$ gives

\[ r=\frac{1\pm\sqrt{1-4}}{2}=\frac{1\pm i\sqrt3}{2}, \] so $r_{2,3}=\dfrac12\pm i\dfrac{\sqrt3}{2}$.

Step 3: Fundamental system

One real root and one complex-conjugate pair $\alpha\pm i\beta$ with $\alpha=\tfrac12$, $\beta=\tfrac{\sqrt3}{2}$ give

\[ y_1=e^{-x}, \qquad y_2=e^{x/2}\cos\frac{\sqrt3}{2}x, \qquad y_3=e^{x/2}\sin\frac{\sqrt3}{2}x. \]

Step 4: Wronskian

The coefficient of $y''$ in the equation is $0$, so by Abel's theorem $W(x)=W(0)$ is a nonzero constant; $y_1,y_2,y_3$ are linearly independent on $\mathbb{R}$.

Step 5: General solution

The solution is recorded in the result below.