Find the general solution of y^((V)) - 10y''' + 9y' = 0

Find the general solution of $y^{(V)} - 10y''' + 9y' = 0$.

Step 1: Characteristic equation

Substituting $y=e^{rx}$: $r^5 - 10r^3 + 9r = 0.$

Step 2: Factor

Factor out $r$, then treat the remaining quartic as a quadratic in $t=r^2$: \[r(r^4-10r^2+9)=0, \qquad t^2-10t+9=(t-1)(t-9)=0 \implies t=1,\ 9.\] So $r^2=1$ or $r^2=9$, giving: $r(r-1)(r+1)(r-3)(r+3)=0.$

Step 3: Roots

Five real, distinct roots: \[r_1=0,\quad r_2=1,\quad r_3=-1,\quad r_4=3,\quad r_5=-3.\]

Step 4: Fundamental system

Each simple real root $r_k$ contributes $e^{r_kx}$: \[y_1=1,\qquad y_2=e^{x},\qquad y_3=e^{-x},\qquad y_4=e^{3x},\qquad y_5=e^{-3x}.\]

Step 5: Wronskian

Writing the equation as $y^{(V)}+p_3y''''+\dots=0$, the coefficient of $y''''$ is $p_3=0$, so by Abel’s theorem $W$ is a nonzero constant. Equivalently, the five exponents $0,1,-1,3,-3$ are distinct, so the Wronskian reduces to a Vandermonde determinant in these exponents, which is nonzero. Hence $y_1,\dots,y_5$ are linearly independent for all $x$.

Step 6: General solution

The solution is recorded in the result below.