Find the general solution of y^((4)) - 5y'' + 4y = 0

Find the general solution of $y^{(4)} - 5y'' + 4y = 0$.

Step 1: Characteristic equation

Substituting $y=e^{rx}$ gives the biquadratic

\[ r^4-5r^2+4=0. \]

Setting $t=r^2$: \[t^2-5t+4=0 \implies (t-1)(t-4)=0 \implies t=1,\ 4.\]

Step 2: Roots

Back-substituting $r^2=t$ gives four real, distinct roots: \[r_1=1, \qquad r_2=-1, \qquad r_3=2, \qquad r_4=-2.\]

Step 3: Fundamental system

\[ y_1=e^x, \qquad y_2=e^{-x}, \qquad y_3=e^{2x}, \qquad y_4=e^{-2x}. \]

Step 4: Wronskian

The coefficient of $y'''$ is $0$, so by Abel's theorem $W(x)$ is constant, equal to its value at $x=0$, $W(0)=72$. Since the four exponents $1,-1,2,-2$ are distinct, this Vandermonde-type determinant is automatically nonzero, confirming $y_1,y_2,y_3,y_4$ are linearly independent.

Step 5: General solution

The solution is recorded in the result below.