Find the general solution of y''' - 8y = 0

Find the general solution of $y''' - 8y = 0$.

Step 1: Characteristic equation

Substituting $y=e^{rx}$: $r^3-8=0.$

This is a difference of cubes, $r^3-2^3=(r-2)(r^2+2r+4)$, confirmed by expansion, so: $(r-2)(r^2+2r+4)=0.$

Step 2: Roots

The first factor gives the real root $r_1=2$. For the second factor, the discriminant is $\Delta=2^2-4(1)(4)=4-16=-12<0$, so: \[r=\frac{-2\pm\sqrt{-12}}{2}=\frac{-2\pm2i\sqrt3}{2}=-1\pm i\sqrt3 \implies r_{2,3}=-1\pm i\sqrt3,\quad \alpha=-1,\ \ \beta=\sqrt3.\]

Step 3: Fundamental system

The real root contributes $e^{2x}$; the complex conjugate pair $r=\alpha\pm i\beta$ contributes $e^{\alpha x}\cos\beta x,\ e^{\alpha x}\sin\beta x$: \[y_1=e^{2x}, \qquad y_2=e^{-x}\cos(\sqrt3\,x), \qquad y_3=e^{-x}\sin(\sqrt3\,x).\]

Step 4: Wronskian

Writing the equation as $y'''+p_2y''+p_1y'+p_0y=0$, the coefficient of $y''$ is $p_2=0$, so by Abel’s theorem $W'=-p_2W=0$ and $W$ is a nonzero constant. Evaluating the $3\times3$ Wronskian determinant at $x=0$ confirms this directly:

\[ W(0)=\begin{vmatrix}1 & 1 & 0\\ 2 & -1 & \sqrt3\\ 4 & -2 & -2\sqrt3\end{vmatrix} = 1\big[(-1)(-2\sqrt3)-\sqrt3(-2)\big]-1\big[2(-2\sqrt3)-\sqrt3(4)\big] = 4\sqrt3+8\sqrt3 = 12\sqrt3 \neq 0, \]

confirming $y_1,y_2,y_3$ are linearly independent for all $x$.

Step 5: General solution

The solution is recorded in the result below.