Find the general solution of y''' - 5y'' + 17y' - 13y = 0

Find the general solution of $y''' - 5y'' + 17y' - 13y = 0$.

Step 1: Characteristic equation

Substituting $y=e^{rx}$ gives

\[ r^3-5r^2+17r-13=0. \]

Testing $r=1$: $1-5+17-13=0$, so $r=1$ is a root, and the cubic factors as

\[ (r-1)(r^2-4r+13)=0. \]

Step 2: Solve for the roots

The linear factor gives $r_1=1$. The quadratic factor $r^2-4r+13=0$ gives

\[ r=\frac{4\pm\sqrt{16-52}}{2}=\frac{4\pm6i}{2}=2\pm3i, \] so $r_{2,3}=2\pm3i$.

Step 3: Fundamental system

\[ y_1=e^x, \qquad y_2=e^{2x}\cos3x, \qquad y_3=e^{2x}\sin3x. \]

Step 4: Wronskian

By Abel's theorem, $W(x)=W(0)e^{5x}$ (the coefficient of $y''$ is $-5$); since $W(0)\neq0$, $W(x)\neq0$ for all $x$, so $y_1,y_2,y_3$ are linearly independent.

Step 5: General solution

The solution is recorded in the result below.