Find the general solution of y^((V)) + 8y''' + 16y' = 0

Find the general solution of $y^{(V)} + 8y''' + 16y' = 0$.

Step 1: Characteristic equation

(The equation is read as $y^{(V)}+8y'''+16y'=0$, the reading that produces a factorable characteristic polynomial.) Substituting $y=e^{rx}$: $r^5 + 8r^3 + 16r = 0.$

Step 2: Factor

Factor out $r$; the remaining quartic is a perfect square in $r^2$: $r(r^4+8r^2+16)=0 \implies r(r^2+4)^2=0.$

Step 3: Roots

One simple real root and a complex conjugate pair of multiplicity two: \[r_1=0 \ \ (\text{simple}), \qquad r=\pm2i \ \ (\text{each of multiplicity } k=2).\]

Step 4: Fundamental system

The simple root contributes $y_1=1$. A complex pair $r=\alpha\pm i\beta$ of multiplicity $k$ contributes $2k$ real solutions of the form $x^j e^{\alpha x}\cos\beta x$, $x^j e^{\alpha x}\sin\beta x$ for $j=0,\dots,k-1$; here $\alpha=0,\ \beta=2,\ k=2$:

\[ y_1=1,\quad y_2=\cos2x,\quad y_3=\sin2x,\quad y_4=x\cos2x,\quad y_5=x\sin2x. \]

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; the root-multiplicity construction guarantees $y_1,\dots,y_5$ are linearly independent.

Step 6: General solution

The solution is recorded in the result below.