Find the general solution of y^((4)) - y''' - y'' - y' + 12y…

Find the general solution of $y^{(4)} - y''' - y'' - y' + 12y = 0$.

Step 1: Characteristic equation

Substituting $y=e^{rx}$ gives

\[ r^4-r^3-r^2-r+12=0. \]

Step 2: No rational root

By the Rational Root Theorem, any rational root must divide $12$, so the candidates are $\pm1,\pm2,\pm3,\pm4,\pm6,\pm12$. Sampling a few, \[r=1:\ 1-1-1-1+12=10\neq0, \qquad r=2:\ 16-8-4-2+12=14\neq0,\]

and direct evaluation at every remaining candidate likewise gives a nonzero value, so the quartic has no rational root.

Step 3: No integer quadratic factorization

Suppose $r^4-r^3-r^2-r+12=(r^2+ar+b)(r^2+cr+d)$ with $b,d$ integers, $bd=12$. Expanding and matching coefficients: \[a+c=-1, \qquad b+d+ac=-1, \qquad ad+bc=-1, \qquad bd=12.\]

From the first equation $c=-1-a$; substituting into the third and using $bd=12$ forces, for each integer factor pair $(b,d)$ of $12$, a specific linear equation in $a$ whose solution is then checked against the second equation $b+d+ac=-1$. Carrying this out for every factor pair $(\pm1,\pm12), (\pm2,\pm6), (\pm3,\pm4)$ (and their swaps) leaves no pair for which $a$ comes out both consistent and rational – each attempt forces $a$ to satisfy two incompatible equations simultaneously. Hence the quartic does not factor into two quadratics with integer (or rational) coefficients; the true factorization requires the irrational coefficients described below, and no clean closed form of the type $(r^2+ar+b)(r^2+cr+d)$ with rational $a,b,c,d$ exists.

Step 4: Roots via the resolvent cubic

Because the quartic is irreducible over the rationals, its exact roots come from the general quartic formula, which for this polynomial routes through an irreducible resolvent cubic (casus irreducibilis) – the roots are genuine real/complex numbers but have no simple closed form in radicals of nice appearance. Solving numerically to high precision and confirming each value by direct back-substitution into $r^4-r^3-r^2-r+12$ (all four give $\approx0$) yields two complex-conjugate pairs:

\[ r_{1,2}=\alpha_1\pm i\beta_1 \approx -1.184608 \pm 1.285993\,i, \qquad r_{3,4}=\alpha_2\pm i\beta_2 \approx 1.684608 \pm 1.042795\,i. \]

(Sanity checks: the sum of all four roots is $2\alpha_1+2\alpha_2\approx2(-1.184608+1.684608)=1.000000$, matching $-(-1)/1$ from the $r^3$ coefficient; the product is $(\alpha_1^2+\beta_1^2)(\alpha_2^2+\beta_2^2)\approx3.05707\times3.92532\approx12.0000$, matching the constant term $12$.)

Step 5: Fundamental system

Two complex-conjugate pairs of roots give

\[ y_1=e^{\alpha_1x}\cos\beta_1x,\quad y_2=e^{\alpha_1x}\sin\beta_1x,\quad y_3=e^{\alpha_2x}\cos\beta_2x,\quad y_4=e^{\alpha_2x}\sin\beta_2x. \]

Step 6: Wronskian

The coefficient of $y'''$ is $-1$, so by Abel's theorem $W(x)=W(0)e^{x}$; since the four roots are distinct, $W(0)\neq0$ and hence $W(x)\neq0$ for all $x$, confirming $y_1,y_2,y_3,y_4$ are linearly independent.

Step 7: General solution

where $\alpha_1\approx-1.184608$, $\beta_1\approx1.285993$, $\alpha_2\approx1.684608$, $\beta_2\approx1.042795$.