Find the general solution of y^((4)) - y = 0

Find the general solution of $y^{(4)} - y = 0$.

Step 1: Characteristic equation

Substituting $y=e^{rx}$ gives

\[ r^4-1=0. \]

Factoring as a difference of squares, twice: $(r^2-1)(r^2+1)=0 \implies (r-1)(r+1)(r^2+1)=0.$

Step 2: Roots

\[ r_1=1, \qquad r_2=-1, \qquad r_{3,4}=\pm i. \]

Step 3: Fundamental system

\[ y_1=e^x, \qquad y_2=e^{-x}, \qquad y_3=\cos x, \qquad y_4=\sin x. \]

Step 4: Wronskian

The coefficient of $y'''$ in the equation is $0$, so by Abel's theorem $W(x)$ is constant; evaluating at $x=0$ gives $W(0)=-8\neq0$, so $y_1,y_2,y_3,y_4$ are linearly independent on $\mathbb{R}$.

Step 5: General solution

The solution is recorded in the result below.